Limit- almost finished, however hint needed. I tried to calculate following limit: (let me present my solution first, and then point errors or give hints how to finish)$$\lim_{n\rightarrow\infty}\frac{(n!)^n}{n^{{n}^{2}}}$$So my partly solution is: Let $a_{n}=\frac{(n!)^n}{n^{{n}^{2}}}$. Then $a_{n}=(\frac{(n!)}{n^{n}})^n$ and denoting $p_{n}=ln (a_{n})$ we have $p_{n}=ln((\frac{(n!)}{n^{n}})^n)$$=n*ln\frac{(n!)}{n^{n}}=$$n*(ln(n!)-n*ln(n)$ and now I use Stirling formula  $n!\approx\sqrt{2\pi n}*(\frac{n}{e})^n$ and now $p_{n}=n*(ln(\sqrt{2\pi n}*(\frac{n}{e})^n)-n*ln(n))$$=n(ln(\sqrt{2\pi n)}+n*ln(n)-n-n*ln(n))=$$=n((ln(\sqrt{2\pi n})+n))=$$=n(ln(n\sqrt{{2n\pi n})})$ and that would imply that $\lim_{n\rightarrow\infty}a_{n}=exp(n\sqrt{2\pi n})^n$ what does not seem good... Where am I doing mistake, is it wrong way of acting? Any hints are desired, thank you in advance.
 A: You can see here that 
$$\lim_{n\to\infty}\frac{n!}{n^n}=0$$
thus
$$(**)\;\;\;\;\;\;\;\;\;\;\;\;\;\sqrt[n]{\frac{(n!)^n}{n^{n^2}}}=\frac{n!}{n^n}\xrightarrow [n\to\infty]{} 0$$
From where it follows that the series
$$\sum_{n=1}^n\frac{(n!)^n}{n^{n^2}}$$
converges, since what we did in (**) above is merely apply Cauchy's Test of the n-th root, and thus the sequence's limit must be zero.
A: There is an error in your third step in the computation of $p_n$.
\begin{align}
p_n & = n (\ln(n!) - n \ln(n))\\
& \sim n (\ln(\sqrt{2 \pi n} \cdot (n/e)^n) - n \ln(n)) & (\because \text{Stirling's formula})\\
& = n (\ln(\sqrt{2 \pi n}) + n \ln n -n - n \ln n)\\
& = \color{blue}{\underbrace{n(\ln(\sqrt{2 \pi n}) - n) \to - \infty}_{n \to \infty}}
\end{align}
Hence, $\ln(a_n) \to -\infty$ as $n \to \infty$. Hence, what can you say about $a_n$?
EDIT
A simpler proof is as follows. Note that $n! \leq \dfrac{n^n}2$ for all $n > 1$. Hence, $$0 < \dfrac{(n!)^n}{n^{n^2}} = \left(\dfrac{n!}{n^n} \right)^n \leq \left(\dfrac12 \right)^n = \dfrac1{2^n}$$ Hence, $$0 \leq \lim_{n \to \infty} \dfrac{(n!)^n}{n^{n^2}} \leq \lim_{n \to \infty} \dfrac1{2^n} = 0$$
