Proof that $n^n<(n!)^2$ for $n>2$ 
Prove that $n^n<(n!)^2$ for $n>2$

I tried math induction, but couldn't prove that $(k+1)^{k+1}<((k+1)!)^2$.

Here is an argument based on Stirling's Approximation:
$$
\frac{n^n}{(n!)^2}\leq \frac{e^n}{n!\sqrt{2\pi n}}\leq \frac{e^3}{6\sqrt{6\pi}}< 0.78.
$$
This follows from the observation that the sequence $\left(\frac{e^n}{n!\sqrt{2\pi n}}\right)_{n\geq 3}$ is decreasing since $n\geq e$ in this range.
