Show that $n^n<(n!)^2$ I want to show that $\lim\limits_{n \to \infty}\frac{n^n}{(n!)^2}=0$
But I have absolutely no idea besides that $\frac{n^n}{(n!)^2}=\frac{n}{1}\cdot \frac{n}{2}\cdot ...\cdot \frac{n}{(n-1)^2}\cdot \frac{n}{n^2}$
Help me please.

Let's check ratio of $a_n$ and $a_{n+1}$:
$$a_n = \frac{n^n}{n!^2}$$
$$a_{n+1} = \frac{(n+1)^{n+1}}{(n+1)!^2}$$
$$\frac{a_{n+1}}{a_n}=\frac{(n+1)^{n+1}}{(n+1)!^2} : \frac{n^n}{n!^2} =
\frac{(n+1)^{n+1}}{n^n}\frac{n!^2}{(n+1)!^2} = \left(1+\frac{1}{n}\right)^n \frac{n!^2(n+1)}{n!^2(n+1)^2} = \frac{1}{n+1}\left(1+\frac{1}{n}\right)^n \sim\frac en,
$$
hence
$$a_n \sim \frac{e^n}{n!}\to 0.$$
