How to show that $\forall n \geq 1$, $\frac{n^{2n}}{n!^2} \geq (\frac{n+1}{n})^{n^2-n}$?

Question

I tried using the fact that $(n+1)/n \leq 2$ and thus $\frac{n^{2n}}{n!^2} \geq 2^{n^2-n}$ but this does not seem to be true.

Answer 1

We can proceed by induction on $n$. The case $n=1$ is easy to check. Suppose that the inequality holds for $1,2,\ldots,n$. Then \begin{align*} \frac{{(n + 1)^{2n + 2} }}{{(n + 1)!^2 }} = \frac{{(n + 1)^{2n} }}{{n!^2 }} &= \frac{{n^{2n} }}{{n!^2 }}\left( {\frac{{n + 1}}{n}} \right)^{2n} \ge \left( {\frac{{n + 1}}{n}} \right)^{n^2 - n} \left( {\frac{{n + 1}}{n}} \right)^{2n} \\& = \left( {\frac{{n + 1}}{n}} \right)^{(n + 1)^2 - (n + 1)} \ge \left( {\frac{{n + 2}}{{n + 1}}} \right)^{(n + 1)^2 - (n + 1)} , \end{align*} which completes the proof.