Factorial Algorithms in different languages

Answer 1

Perl6

sub factorial ($n) { [*] 1..$n }

I hardly know about Perl6. But I guess this [*] operator is same as Haskell's product.

This code runs on Pugs, and maybe Parrot (I didn't check it.)

Edit

This code also works.

sub postfix: ($n) { [*] 1..$n }

# This function(?) call like below ... It looks like mathematical notation.
say 10!;