Sorry for the necropost, but I just read this and the answer came to me while cooking lunch.
Not that hard to show this is not possible: first, check this does not hold for 24!. Then notice that if n>24, then n>10. This is obvious, but notice that multiplying by anything larger than or equal to 10, the factorial will gain at least one digit in length. So , Let N(k!) be the number of digits in k! . We are trying to see when/if we can have N(k!)=k.
Then, given 23! has 23 digits , i.e., N(23!)=23, and N(24!)>24 , we see that N(k!) will increase at least by 1 from k to (k+1), while the digits increase just by 1 , i.e., when we go from N(24!) to N(25!) , k goes from 24 to 25, i.e., increases precisely by 1, while k, the number of digits, increases at least by 1, so there is no way N(k!) can catch up with k , since N(24!)> 24, and the expression in the left will increase at least as fast as the expression in the right. I leave it to you to see what happens when we use a number base different from decimal.
So you have two expressions A,B ; A>B for n>24, and then A goes on to increase faster than B after 24, so B cannot catch up with A.
. People may just be curious only not really interested or fall for them. 