Compute $\lim_{n\rightarrow\infty}\frac{n^n}{(n!)^2}$ I have to compute $\lim_{n\rightarrow\infty}\frac{n^n}{(n!)^2}$. 
I tried say that this limit exists and it's l, so we have $\lim_{n\rightarrow\infty}\frac{n^n}{(n!)^2} = L$ then I rewrited it as:
$\lim_{n\rightarrow\infty}(\frac{\sqrt n}{\sqrt[n]{n!}})^{2n}$ then I used natural log over the whole expresion but didn't got into a nice place.
I don't know about Pi function or gamma function so therefore can't really use L'Hospital's rule.
 A: Another way. Note that as $n\to +\infty$,
$$\frac{n^n}{(n!)^2}=\prod_{k=1}^n\frac{n}{ k(n+1-k)}\leq \frac{n}{ \lfloor\frac{n+1}{2}\rfloor (n+1-\lfloor\frac{n+1}{2}\rfloor)}\leq \frac{4}{n}\to 0$$
because each factor $\frac{n}{ k(n+1-k)}\leq 1$, and the smallest one is in the middle.
A: By ratio test
$$\frac{(n+1)^{n+1}}{((n+1)!)^2}\frac{(n!)^2}{n^n}=\frac1{n+1}\left(1+\frac1n\right)^n\to 0$$
then 
$$\lim_{n\rightarrow\infty}\frac{n^n}{(n!)^2}=0$$
A: By the root test, the limit is
$n/(n!)^{2/n}$.
By looking at the last 2/3 of 1 to n,
$n! > (n/3)^{2n/3}$
so $(n!)^{2/n} > (n/3)^{4/3}$
so the  ratio is less than
$3^{4/3}/n^{1/3}$
which goes to zero.
This method csn be used to shiw that
$n^n/(n!)^a \to 0$
for any $a > 1$.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

With $\ds{n!}$-Stirling Asymptotic Expansion:

\begin{align}
\lim_{n \to \infty}{n^{n} \over \pars{n!}^{2}} & =
\lim_{n \to \infty}{n^{n} \over \bracks{\root{2\pi}n^{n + 1/2}\expo{-n}}^{2}} =
{1 \over 2\pi}\lim_{n \to \infty}\bracks{{1 \over n}\pars{\expo{2} \over n}^{n}} \\[5mm] & =
{1 \over 2\pi}\lim_{n \to \infty}{\exp\pars{n\bracks{2 - \ln\pars{n}}} \over n} = \bbx{\large 0}
\end{align}
