Joe, 5 3 bytes (+2 or +3 for -t flag)

Well, apparently I didn't utilize the full potential of Joe. This was possible back when I first posted this.

\AR

Here, R gives the range from 0 to n, exclusive. Then \A takes successive prefixes of it (A is the identity function). Examples:

With -t flag (note: this is now the standard output even without the flag):

   (\AR)5
0
0 1
0 1 2
0 1 2 3
0 1 2 3 4
   \AR5
0
0 1
0 1 2
0 1 2 3
0 1 2 3 4
   \AR2
0
0 1
   \AR1
0
   \AR0

Without it:

   \AR5
[[0], [0, 1], [0, 1, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4]]
   (\AR)5
[[0], [0, 1], [0, 1, 2], [0, 1, 2, 3], [0, 1, 2, 3, 4]]
   \AR2
[[0], [0, 1]]
   \AR1
[[0]]
   \AR0
[]

The rules got changed a bit. My old code didn't behave correctly with N =0. Also, now output could be just a nested list, so -t can be dropped.

1R1+R

Now, Rn gives a range from 0 to n, exclusive. If given 0, it returns an empty list. 1+ adds 1 to every element of that range. 1R maps the values to ranges from 1 to x. Empty liats, when mapped, return empty lists.

Example output:

   1R1+R0
[]
   1R1+R5
[[1], [1, 2], [1, 2, 3], [1, 2, 3, 4], [1, 2, 3, 4, 5]]

Update: I just noticed something. The function automatically maps to rank 0 elements. The following example is run with -t flag.

   1R1+R3 5 8
1
1 2
1 2 3

1
1 2
1 2 3
1 2 3 4
1 2 3 4 5

1
1 2
1 2 3
1 2 3 4
1 2 3 4 5
1 2 3 4 5 6
1 2 3 4 5 6 7
1 2 3 4 5 6 7 8

Old: 5 bytes (with the -t flag)

1R1R

This is an anonymous function which takes in a number, creates a list from 1 to N (1Rn) and maps those values to the preceding range, giving a range from 1 to x for each item of range 1 to N.

The -t flag gives output as a J-like table.

   1R1R5
1
1 2
1 2 3
1 2 3 4
1 2 3 4 5

Note : the language is very new and not complete, but the latest version was released before this challenge.

Python 3, 48 45 bytes

f=lambda n:n and[f(n-1),print(*range(1,n+1))]

Hooray for side effects.

APL, 5

⍪⍳¨⍳⎕

creates a vector 1..n and for each element another such vector.

Then ⍪ makes a column out of all vectors. This avoids the problem with trailing blanks.

Try it on tryapl.org

Older solution:

{⎕←⍳⍵}¨⍳⎕

Creates a vector 1..n

{⎕←⍳⍵} is a function that outputs for each (¨) element a vector 1..n on a separate line

This one can't be tried on tryapl.org unfortunately, because ⎕← doesn't work there.

J, 27 bytes

J is not good with non-array numeric output. This function creates a properly formatted string from the numbers.

   ;@(<@,&LF@":@:>:@:i.@>:@i.)

   (;@(<@,&LF@":@:>:@:i.@>:@i.)) 4
1
1 2
1 2 3
1 2 3 4

Try it online here.

PHP, 53 Bytes

Edit 2: Ismael Miguel suggested reading from input instead of defining a function, so the score is now 53 bytes for PHP:

for($a=1;@$i++<$n=$argv[1];$a.=" ".($i+print"$a\n"));

And once again, it can be improved if PHP is configured to ignore errors (52 bytes):

for($a=1;$i++<$n=$argv[1];$a.=" ".($i+print"$a\n"));
for($a=1;$i++<$n=$_GET[n];$a.=" ".($i+print"$a\n"));

Edit: Austin suggested a 60 bytes version in the comments:

function f($n){for($a=1;@$i++<$n;$a.=" ".($i+print"$a\n"));}

Which can be improved if we doesn't display PHP errors (59 bytes):

function f($n){for($a=1;$i++<$n;$a.=" ".($i+print"$a\n"));}

$a stores the next line that will be printed, and each time it's printed a space and the next number (print always returns 1) are concatened to it.

Recursive functions (65 bytes):

function f($n){$n>1&&f($n-1);echo implode(' ',range(1,$n))."\n";}
function f($n){$n>1&&f($n-1);for(;@$i++<$n;)echo$i,' ';echo"\n";}   // Using @ to hide notices.

Shorter recursive function, with error reporting disabled (64 bytes):

function f($n){$n>1&&f($n-1);for(;$i++<$n;)echo$i,' ';echo"\n";}

Even shorter recursive function, with error reporting disabled and an empty line before real output (62 bytes):

function f($n){$n&&f($n-1);for(;$i++<$n;)echo$i,' ';echo"\n";}

Just for fun, non-recursive fucntions:

function f($n){for($i=0;$i<$n;print implode(' ',range(1,++$i))."\n");}    // 70 bytes
function f($n){for(;@$i<$n;print implode(' ',range(1,@++$i))."\n");}      // 68 bytes, hiding notices.
function f($n){for(;$i<$n;print implode(' ',range(1,++$i))."\n");}        // 66 bytes, error reporting disabled.

CJam, 13 12 bytes

ri{),:)S*N}/

How it works:

ri{       }/     "Run the block input number of times with iteration index from 0 to N-1";
   )             "Increment the iteration index (making it 1 to N)";
    ,            "Get an array of 0 to iteration index";
     :)          "Increment each of the above array members by 1";
       S*        "Join all above array numbers with space";
         N       "Add a new line. After all iterations, things are automatically printed";

Try it online here

Pyth, 9 bytes

VQjdr1hhN

Really thought that this can be done shorter, but it doesn't seem so.

Try it online.

            Q = input()
VQ          For N in [0, 1, ..., Q-1]:
    r1hhN       create list [1, ..., N+1+1-1]
  jd            print joined with spaces

JavaScript (ES6) 49 52

Such a simple task, I wonder if this can be made shorter in JS (Update: yes, using recursion)

Recursive 49

f=n=>alert((r=w=>n-i++?w+'\n'+r(w+' '+i):w)(i=1))

Iteraive 52

f=n=>{for(o=r=i=1;i++<n;o+='\n'+r)r+=' '+i;alert(o)}

Java, 85 84 bytes

This is surprisingly short in Java.

void a(int a){String b="";for(int c=0;c++<a;System.out.println(b+=(c>1?" ":"")+c));}

Indented:

void a(int a){
    String b="";
    for(int c=0;
        c++<a;
        System.out.println(
                b+=(c>1?" ":"")+c
        ));
}

1 byte thanks to Bigtoes/Geobits

Prolog - 119

h(N):-setof(X,(between(1,N,K),setof(Y,between(1,K,Y),X)),[L]),k(L),nl,fail.
k([A|B]):-write(A),(B=[];write(" "),k(B)).

Python 2 - 62 54 65 bytes

def f(n):
 for x in range(n):print' '.join(map(str,range(1,x+2)))

J, 9 characters

As a tacit, monadic verb.

[:":\1+i.

• i. y – the numbers from 0 to y - 1.
• 1 + i. y – the numbers from 1 to y.
• ": y – the vector y represented as a string.
• ":\ y – each prefix of y represented as a string.
• ":\ 1 + i. y – each prefix of the numbers from 1 to y represented as a matrix of characters.

><> (Fish), 40 37 + 3 = 40 bytes

&1>:&:&)?;1\
(?v:n" "o1+>}:{:@
ao\~1+

Once again, ><> does decently well at another number printing exercise. Run with the -v flag for input, e.g.

py -3 fish.py -v 4

Explanation

&               Put n in register
1               Push 1 (call this "i")

[outer loop]

:&:&)?          If i > n...
;                 Halt
1                 Else push 1 (call this "j")

[inner loop]

}:{:@(?         If j > i...
~1+ao             Pop j, print newline, increment i and go to start of outer loop
:n" "o1+          Else print j, print a space, increment j and go to start of inner loop

Python 2 - 72

>>> def p(N):print'\n'.join(' '.join(map(str,range(1,i+2)))for i in range(N))
... 
>>> p(5)
1
1 2
1 2 3
1 2 3 4
1 2 3 4 5

Perl, 28

Reads the parameter from stdin.

@x=1..$_,print"@x
"for 1..<>

From the command line:

perl -E'$,=$";say 1..$_ for 1..<>'

but I don't now how to count that (probably between 25 and 29).

C (with no loops, yeah!) - 72 bytes

b(n,c){if(n){b(n-1,32);printf("%d%c",n,c);}}r(n){if(n){r(n-1);b(n,10);}}

This creates a function r(n) that can be used this way:

main(){ r(5); }

See it in action, here on tutorialspoint.com

It requires a very few tricks easily explained. I think it can be greatly improved.

Thunno 2 N, 5 bytes

RRıðj

Attempt This Online!

Explanation

RRıðj  # Implicit input    ->  4
R      # Range [1..n]      ->  [1,2,3,4]
 R     # Range of each     ->  [[1],[1,2],[1,2,3],[1,2,3,4]]
  ı    # To each list:
   ðj  #  Join by spaces   ->  ["1","1 2","1 2 3","1 2 3 4"]
       # Join by newlines  ->  "1\n1 2\n1 2 3\n1 2 3 4"
       # Implicit output

Julia 1.0, 34 bytes

~n=1:n.|>x->println(join(1:x,' '))

Try it online!

Python

import string
N,s=int(input()),list(string.digits)
for i in range(1,N+1):
    print(' '.join(s[1:i+1]))

Golfscript 14

,{2+,1>' '*n}/

Expects the input number to be present on the stack.

Online example: link

Clip, 16

Jm[ijkw,1iwS},1n

Explanation

J                   .- join with newlines                           -.
 m[i        },1n    .- map numbers from 1 to numeric value of input -.
    jkw   wS        .- join with spaces                             -.
       ,1i          .- numbers from 1 to index                      -.

Go, 93 81 78 93 90 bytes

func r(n int)(s string){s=string(n+48);if n!=1{s=r(n-1)+" "+s};println(s);return}

Current ungolfed

func r(n int) (s string) {
    // Convert n to a string, we do not have to initialize s since
    // we hijacked the return value.
    // Numbers in the ascii table starts at 48
    s = string(n | 48)
    // Unless we are on our last iteration, we need previous iterations,
    // a space and our current iteration
    if n != 1 {
        // Collect the result of previous iteration for output
        s = r(n-1) + " " + s
    }
    println(s)
    // We can use a naked return since we specified the
    // name of our return value in the function signature
    return
}

If we need to handle N > 9 we can use the following at 78 bytes, however it requires importing the fmt package.

func r(n int)(s string){s=Sprint(n);if n!=1{s=r(n-1)+" "+s};Println(s);return}

If we include the import statement I'm now back at my initial 93 92 90 bytes

import."fmt";func r(n int)(s string){s=Sprint(n);if n>1{s=r(n-1)+" "+s};Println(s);return}

Test it online here: http://play.golang.org/p/BWLQ9R6ilw

The version with fmt is here: http://play.golang.org/p/hQEkLvpiqt

ZX / Sinclair BASIC - 39 bytes

ZX Basic uses 1 byte per keyword (all the uppercase words), so helps to keep the byte size down a bit...

1 INPUT n:FOR i=1 TO n:FOR j=1 TO i:PRINT j;" ";:NEXT j:PRINT:NEXT i

Using n = 8

R, 28

for(i in 1:scan())print(1:i)

TI-Basic, 28 bytes

Input N
For(I,1,N
randIntNoRep(1,N->L1
SortA(L1
Disp L1
End

C, 89 characters

// 90 characters
f(int n){int a=1,b;for(;n--;++a){for(b=0;b<a;++b)printf("%c%d",(!!b)*' ',b+1);puts("");}}

To eliminate confusion about puts("");. This simply prints a newline character (as seen here):

Notice that puts not only differs from fputs in that it uses stdout as destination, but it also appends a newline character at the end automatically (which fputs does not).

I got it slightly shorter with @TheBestOne's java algorithm:

// 89 characters
f(int a){char b[999]="",*p=b+1;int c=0;for(;a--&&(sprintf(b,"%s %d",b,++c)&&puts(p)););}

Perl: 34 characters

print"@$_\n"for map[1..$_],1..$_;

This code gets the input number supplied through the special variable $_.

C# - 94 bytes

Written as an anonymous function that returns a string, which doesn't seem to be disallowed by the spec.

n=>String.Join("\n\n",Enumerable.Range(1,n).Select(l=>String.Join(" ",Enumerable.Range(1,l))))

Here's an ungolfed version (comments are read in BDCA order):

n =>
    String.Join("\n\n",                    //...then join it together with newlines.
        Enumerable.Range(1, n).Select(l => //For each l from 1 to n, ...
                String.Join(" ",              //...and join it with spaces, ...
                    Enumerable.Range(1, l)    //...get the range from 1 to l, ...

Bash+coreutils, 26 bytes

seq $1|sed "x;G;s/\n/ /;h"

• seq simply generates the numbers 1 to n
• sed saves the entire output for a given line in the hold space, and then appends the next line to it.

Haskell, 62 57 bytes

e=enumFromTo 1
f=putStr.unlines.map(unwords.map show.e).e

Point-free style. Usage example:

Prelude> f 5
1
1 2
1 2 3
1 2 3 4
1 2 3 4 5