Inequality Problem of examination

Question

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.

Answer 1

For $n=k, (k!)^2>k^k$

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

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

But $\left(1+\frac1k\right)^k<3$ See this and this

Answer 2

$$(n!)^2=1(n)\times2(n-1)\times3(n-2)\times \cdots n(1)$$ 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 $(n!)^2>n^n$.