As written, clearly the first is bigger because $2012!>2013$, so that $\sqrt[2013]{2013}<\sqrt[2012]{2013}<\sqrt[2012]{2012!}$. I think you wrote it wrong, though. What did you actually mean?

sorry for mistake it must be $\sqrt[2013]{2013!}$

Solution in albania http://prntscr.com/uwai4

What we wish to see is: $\sqrt[2012]{2012!} \Box \sqrt[2013]{2013!}$ Raising both sides to the $2012*2013$ power gives us: $\ 2012!^{2013} \Box 2013!^{2012}$ Now notice that $2012!^{2013} = 2012!^{2012}*2012!$ and that $2013!^{2012} = 2012!^{2012}*2013^{2012}$ Thus we can rewrite our inequality as: $2012!^{2012}*2012! \Box 2012!^{2012}*2013^{2012}$ Dividing both sides by $2012!^{2012}$ reduces the inequality to: $2012! \Box 2013^{2012}$ Of which the RHS is greater (each element in the product of the RHS is greater than any element in the product of the LHS and they have the same number of elements.) So we see that $\sqrt[2012]{2012!} < \sqrt[2013]{2013!}$

Let $x= \sqrt [2012] {2012!}$, then $\newline x^{2012}=2012! \implies x < 2013 \implies x^{2013} < 2012! \cdot 2013= 2013! \implies x < \sqrt [2013] {2013!}$

Generalized: $\forall n \in \mathbb{N}$ we have the following : $(n+1)!^n > n!^{n+1}$

Proof