# How can I show that $n^{n+2}<(2n)!$ for any integer $n$.

When I was try to show that the series $$\sum_{n\in\mathbb{N}} \frac{n^n}{(2n)!}$$ is convergent using comparison test, I stuck at the point $$n^{n+2}<(2n)!$$ I think it can be show using mathematical induction.
Note that $$(n!)^2 = \prod_{i=1}^n i(n+1-i) \ge \prod_{i=1}^n n = n^n$$ Hence $n! \ge n^{n/2}$. This gives $$(2n)! \ge (2n)^n = 2^n n^n$$ For $n \ge 2$ we have $2^n \ge n^2$, hence $(2n)!\ge n^{n+2}$.
Show that $\frac{a_{n+1}}{a_n}$ converge to some value $>1$, keep in mind the definition of $e = \lim_{n \to \infty} (1+\frac{1}{n})^n$. Hence the series diverges.