How to solve $ \lim_{n\to\infty} \frac {(n!)^\frac{1}{n}}{n} $? I need to find the limit for: 
$ \lim_{n\to\infty} \frac {(n!)^\frac{1}{n}}{n} $
I know the answer is $\frac {1}{e}$ but I have no idea how to get that answer.
I'd appreciate some help. 
 A: Of course, there is a way using Stirling's approximation. If you wish to look at some different method:
$\lim\limits_{n\to\infty} \frac {(n!)^\frac{1}{n}}{n}=\exp \left[\lim\limits_{n\to \infty}\frac{1}{n}\sum_{k=1}^n\log \frac{k}{n}\right]=\exp\left[\int_0^1\log x dx\right]=e^{-1}$
A: Let $a_n=\dfrac{n!}{n^n}$. This is a sequence of positive numbers. A well known theorem says that in this case, if $\ell =\lim \dfrac{a_{n+1}}{a_n}$ exists then so does $\ell'=\lim a_n^{1/n}$ and both are equal. You should be able to obtain what $\ell$ is.

One can shed some light on Kunnysan's solution. 

If $f:[a,b]\to\Bbb R$ is properly integrable on $[a,b]$; define $$f_{kn}=f(a+k\delta_n)\; ;\;\delta_n=\frac{b-a}n$$
Then $$\lim\limits_{n\to\infty}\left(f_{1n}f_{2n}\cdots f_{nn}\right)^{1/n}\to \exp\left(\frac{1}{b-a}\int_a^b \log f(x) dx\right)$$
provided $\sup\limits_{[a,b]} f>0$. 

In the case of $f(x)=x$, the result holds, but rather because 

If $f$ is monotone on $(0,1)$ and the (improper) integral $\int_0^1 f$ exists, then $$\frac 1n \sum_{k=1}^{n-1} f\left(\frac{k}n\right)\to\int_0^1 f$$

A: $$\ln  \frac {(n!)^\frac{1}{n}}{n} =\frac{\ln(n!)-\ln n^n}{n}$$
By Stolz Cezaro
$$\lim_n \ln  \frac {(n!)^\frac{1}{n}}{n} =\lim_n \ln((n+1)!)-\ln (n+1)^{n+1} -\ln(n!)+\ln(n^n)= \lim_n \ln \frac{(n+1)!n^n}{n!(n+1)^{n+1}}$$
$$ =\lim_n \ln \frac{n^n}{(n+1)^{n}}=-\lim_n \ln \frac{(n+1)^{n}}{n^n}=-\lim_n \ln \left( 1+\frac{1}{n}\right)^n=-1$$
