Inequality Problem of examination Show that if $n > 2$, then $(n!)^2 > n^n$.
I cannot find any way out to these.Please help.
I tried break $n!$ and then argue but failed.

$$(n!)^2=1(n)\times2(n-1)\times3(n-2)\times \cdots n(1)$$
But
$$n\ge a+1 \implies a(n)\ge a(a+1)\implies (a+1)(n-a)\ge n$$
And equality is achieved iff $a=0$ or $a=n-1$. Since for $n>2$ we have $a=1\neq n-1$ in our product, we have the strict inequality $(n!)^2>n^n$.
