How to find $\lim _{ n\to \infty } \frac { ({ n!) }^{ 1\over n } }{ n } $? How to find $\lim _{ n\to \infty  } \frac { ({ n!) }^{ 1\over n } }{ n } $ ?
I tried taking using logarithm to bring the expression to sum form and then tried L Hospital's Rule.But its not working.Please help!
This is what wolfram alpha is showing,but its not providing the steps!
BTW if someone can tell me a method without using integration, I'd love to know!

Note
\begin{align}\frac{(n!)^{1/n}}{n} &= \left[\left(1 - \frac{0}{n}\right)\left(1 - \frac{1}{n}\right)\left(1 - \frac{2}{n}\right)\cdots \left(1 - \frac{n-1}{n}\right)\right]^{1/n}\\
&= \exp\left\{\frac{1}{n}\sum_{k = 0}^{n-1} \log\left(1 - \frac{k}{n}\right)\right\}
\end{align}
and the last expression converges to 
$$\exp\left\{\int_0^1\log(1 - x)\, dx\right\} = \exp(-1) = \frac{1}{e}.$$
Alternative: If you want to avoid integration, consider the fact that if $\{a_n\}$ is a sequence of positive real numbers such that $\lim\limits_{n\to \infty} \frac{a_{n+1}}{a_n} = L$, then $\lim\limits_{n\to \infty} a_n^{1/n} = L$.
Now $\frac{(n!)^{1/n}}{n} = a_n^{1/n}$, where $a_n = \frac{n!}{n^n}$. So 
$$\frac{a_{n+1}}{a_n} = \frac{(n+1)!}{(n+1)^{n+1}}\cdot \frac{n^n}{n!} = \frac{n+1}{n+1}\cdot\frac{n^n}{(n+1)^n} = \left(\frac{n}{n+1}\right)^n = \left(\frac{1}{1 + \frac{1}{n}}\right)^n = \frac{1}{\left(1 + \frac{1}{n}\right)^n}.$$
Since $\lim\limits_{n\to \infty} (1 + \frac{1}{n})^n = e$, then $$\lim_{n\to \infty} \frac{a_{n+1}}{a_n} = \frac{1}{e}.$$
Therefore $$\lim_{n\to \infty} \frac{(n!)^{1/n}}{n} = \frac{1}{e}.$$
