Find $\lim_{n\to\infty} \frac{(n!)^{1/n}}{n}$. 
Find $$\lim_{n\to\infty} \frac{(n!)^{1/n}}{n}.$$

I don't know how to start. Hints are also appreciated.

Let $f_n=\frac{n!}{n^n}$ so that $f_{n+1}= \frac{(n+1)!}{(n+!)^n+1}$
So $\frac{f_{n+1}}{f_n}=(1+\frac{1}{n})^{-n}$
Therefore 
$\lim_{n\to\infty}(1+\frac{1}{n})^{-n}=\frac{1}{e}>0 $
We know that if $<f_n>$ is a sequence such that $f_n$ is greater than 0 for all n, and $\lim_{n\to\infty}\frac{f_{n+1}}{f_n}=l, l>0$. Then $\lim_{n\to\infty}{f_n}^{\frac{1}{n}}=l$.
Using the above theorem we have $\lim_{n\to\infty} \frac{(n!)^{\frac{1}{n}}}{n}=\frac{1}{e}$
