Show $\left( n!\right)^2 > n^n$.

We do the induction step. If $(k!)^2\gt k^k$, and $k\ge 3$, we show that $((k+1)!)^2 \gt (k+1)^{k+1}$, Note that $((k+1)!)^2 =(k+1)^2(k!)^2 = (k+1)^2(k!)^2 \gt (k+1)^2 k^k$.

We show that $(k+1)^2 k^k \gt (k+1)^{k+1}$, or equivalently that $(k+1)k^k \gt (k+1)^k$. So we need to show that $k+1 \gt \left(1+\frac{1}{k}\right)^k$. This is true, since $\left(1+\frac{1}{k}\right)^k \lt e$.

Another way: Note that $(n!)^2 =\prod_1^n k(n+1-k)$. The function $x(n+1-x)$ increases until $x=\frac{n+1}{2}$, then decreases. In particular, $k(n+1-k)\ge (1)(n)=n$. So the product $\prod_1^n k(n+1-k)$ is $\ge$ the product of $n$ copies of $n$.

Base case, $n=3$, is true by $(3!)^2 = 36 > 27 = 3^3$.

Suppose that we have for $k>2$, $(k!)^2 > k^k$ (IH), now we want to show that $((k+1)!)^2 > (k+1)^{k+1}$.

\begin{align*} ((k+1)!)^2 &= ((k+1)(k!))^2\\ &=(k+1)^2(k!)^2>(k+1)^2k^k \qquad\text{ (where the inequality is from IH)} \end{align*}

So if we can show that $(k+1)^2k^k>(k+1)^{k+1}$ then we are done. I'll leave that up to you.