Showing that $\frac{\sqrt[n]{n!}}{n}$ $\rightarrow \frac{1}{e}$

Have not found a way to rewrite your expression to get the desired result. However, here is a suggested approach.

Maybe rewrite the left-hand side as $$\sqrt[n]{\frac{n!}{n^n}}.$$

Take the logarithm. We get $$\frac{1}{n}\left(\log\left(\frac{1}{n}\right)+ \log\left(\frac{2}{n}\right)+\log\left(\frac{3}{n}\right)+\cdots+\log\left(\frac{n}{n}\right)\right).$$

Now think of the above sum as a Riemann sum for the not quite proper integral $$\int_0^1 \log x\,dx.$$

I would like to use the following lemma:

If $\lim_{n\to\infty}a_n=a$ and $a_n>0$ for all $n$, then we have $$ \lim_{n\to\infty}\sqrt[n]{a_1a_2\cdots a_n}=a \tag{1} $$

Let $a_n=(1+\frac{1}{n})^n$, then $a_n>0$ for all $n$ and $\lim_{n\to\infty}a_n=e$. Applying ($*$) we have $$ \begin{align} e&=\lim_{n\to\infty}\sqrt[n]{a_1a_2\cdots a_n}\\ &=\lim_{n\to\infty}\sqrt[n]{\left(\frac{2}{1}\right)^1\left(\frac{3}{2}\right)^2\cdots\left(\frac{n+1}{n}\right)^n}\\ &=\lim_{n\to\infty}\sqrt[n]{\frac{(n+1)^n}{n!}}\\&= \lim_{n\to\infty}\frac{n+1}{\sqrt[n]{n!}}=\lim_{n\to\infty}\frac{n}{\sqrt[n]{n!}}+\lim_{n\to\infty}\frac{1}{\sqrt[n]{n!}}=\lim_{n\to\infty}\frac{n}{\sqrt[n]{n!}} \end{align}\tag{2} $$ where we use (1) in the last equality to show that $ \lim_{n\to\infty}\frac{1}{\sqrt[n]{n!}}=0. $

It follows from (2) that $$ \lim_{n\to\infty}\frac{\sqrt[n]{n!}}{n}=\frac{1}{e}. $$

If $a_n \geq 0$, then the following inequality holds:

$$ \liminf_{n\to\infty} \frac{a_{n+1}}{a_n} \leq \liminf_{n\to\infty} \sqrt[n]{a_n} \leq \limsup_{n\to\infty} \sqrt[n]{a_n} \leq \limsup_{n\to\infty} \frac{a_{n+1}}{a_n}. $$

Now let $ a_n = n! / n^n $. Then it follows that

$$ \frac{a_{n+1}}{a_n} = \frac{\frac{(n+1)!}{(n+1)^{n+1}}}{\frac{n!}{n^n}} = \frac{1}{\left(1+\frac{1}{n}\right)^n},$$

and hence

$$ \liminf_{n\to\infty} \frac{a_{n+1}}{a_n} = \limsup_{n\to\infty} \frac{a_{n+1}}{a_n} = \frac{1}{e}. $$

This proves that $\sqrt[n]{a_n} \to e^{-1}$ .