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.
 A: 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
$$a_n \sim \frac{e^n}{n!}\to 0.$$
A: Taking logarithms and using formula for arithmetic sum: $$n \log(n) - 2 \sum_{i=1}^n i = n\log(n) - 2n(n-1)/2 = n(\log(n) - n-1)$$This obviously goes to $-\infty$ since any polynomial dominates a logarithm and if the logarithm of something goes to $-\infty$, itself it must go to 0.
A: Let $a_{n}=\frac{n^{n}}{\left( n!\right) ^{2}}.$ We compute , as Michael
Galuza did
\begin{equation*}
\frac{a_{n+1}}{a_{n}}=\frac{(1+\frac{1}{n})^{n}}{n+1}.
\end{equation*}
Note that for large $n,$ 
\begin{equation*}
\frac{a_{n+1}}{a_{n}}<1,
\end{equation*} so the positive sequence $(a_{n})$ is decreasing and bounded from below by 
$0.$ Then there exists $l\geq 0$ such that $\lim_{n\rightarrow \infty
}a_{n}=l.$ ($(a_{n})$ converges to $l$). Assume that $l>0.$ Since $(a_{n+1})$
is a subsequence of $($a$_{n})$ it converges to $l$ too. Then
\begin{equation*}
\lim_{n\rightarrow \infty }\frac{a_{n+1}}{a_{n}}=\frac{l}{l}=1.
\end{equation*}
However
\begin{equation*}
\lim_{n\rightarrow \infty }\frac{a_{n+1}}{a_{n}}=\lim_{n\rightarrow +\infty }%
\frac{1}{n+1}(1+\frac{1}{n})^{n}=0\cdot e=0.
\end{equation*}
