Show that this limit is related to Euler number I am calculating the limit $\lim_{n \rightarrow \infty} \left( \frac{n!^{\frac{1}{n}}}{n} \right)= \frac{1}{e}.$ I got this limit from wolframalpha, but don't know how to show this.wolframalpha

You do not need Stirling's approximation or logarithms. Since:
$$ n=\prod_{k=1}^{n-1}\left(1+\frac{1}{k}\right) $$
we have:
$$ N! = \prod_{n=2}^{N}\prod_{k=1}^{n-1}\left(1+\frac{1}{k}\right) = \prod_{k=1}^{N-1}\left(1+\frac{1}{k}\right)^{N-k}=\frac{N^N}{\prod_{k=1}^{N-1}\left(1+\frac{1}{k}\right)^k}$$
so, if we define:
$$ a_N = \frac{N!}{N^N} $$
we have:
$$ \frac{a_{N+1}}{a_N}=\left(1+\frac{1}{N}\right)^{-N}$$
so:
$$ \lim_{N\to +\infty}\frac{a_{N+1}}{a_N}=\frac{1}{e}$$
implies:
$$ \lim_{N\to +\infty}\sqrt[N]{a_N}=\frac{1}{e}$$
as wanted.
