Proving inequality $3^{n^2} > (n!)^4$ Prove that $3^{n^2} > (n!)^4$ for all positive integers $n$. 
I tried to use induction on this problem but failed to do so. I instead tried to prove $3^{2n+1}>(n+1)^4$, but couldn't come up with the solution.

Clearly, $n^n > n!$ for $n>1$ and so $n^{4n} > (n!)^4$. Therefore, we only need to prove that $3^{n^2} \ge n^{4n}$, or equivalently, $3^n \ge n^4$. This is true for $n \ge 8$. For $n <8$, we just verify explicitly that $3^{n^2} > (n!)^4$.
