real analysis - Showing that $frac{sqrt[n]{n!}}{n}$ $rightarrow frac{1}{e}$

Show:

$$\lim_{n\to\infty}\frac{\sqrt[n]{n!}}{n}= \frac{1}{e}$$

So I can expand the numerator by geometric mean. Letting $C_{n}=\left(\ln(a_{1})+...+\ln(a_{n})\right)/n$. Let the numerator be called $a_{n}$ and the denominator be $b_{n}$ Is there a way to use this statement so that I could force the original sequence into the form of $1/\left(1+\frac{1}{n}\right)^n$

Answer

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

$$\lim_{n\to\infty}\frac{\sqrt[n]{n!}}{n}=\frac{1}{e}.$$