Inequality Problem of examination Show that if $n &gt; 2$, then $(n!)^2 &gt; n^n$.

I cannot find any way out to these.Please help.

I tried break $n!$ and then argue but failed.
For $n=k, (k!)^2&gt;k^k$

For $n=k+1,\{(k+)!\}^2=(k+1)^2\cdot(k!)^2&gt;(k+1)^2\cdot k^k$

It is sufficient to prove $(k+1)^2\cdot k^k&gt;(k+1)^{k+1}\iff k+1&gt;\left(1+\frac1k\right)^k$

But $\left(1+\frac1k\right)^k&lt;3$ See this and this
$$(n!)^2=1(n)\times2(n-1)\times3(n-2)\times \cdots n(1)$$
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 $n&gt;2$ we have $a=1\neq n-1$ in our product, we have the strict inequality $(n!)^2&gt;n^n$.