Show that $\displaystyle\lim_{n\to\infty}{\frac{n}{\sqrt[n]{n!}}}=e$ Here I want to show that 
$$\lim_{n\to\infty}{\frac{n}{\sqrt[n]{n!}}}=e.$$
Based on these questions, Question 1 and Question 2, we have 
$$\frac{n}{\sqrt[n]{n!}}=\sqrt[n]{\frac{n^n}{n!}}\leq\sqrt[n]{e^{n-1}}<e.$$
But I did not find any clue on the lower bound if I want to use the sandwich principle. 
My intuition is to find an expression, say $f(n)\leq n^n/n!$, such that $f(n)^{1/n}$ would converge to $e$. But this somewhat seems very difficult. I know the result is almost immediate if we use Stirling's formula, but I still want to seek for an elementary proof to this limit. 
 A: It suffices to show that
$$
\ln\frac{(n!)^{\frac{1}{n}}}{n}=\ln\left(\frac{n!}{n^n}\right)^{\frac{1}{n}}=\ln\left(\prod_{k=1}^n\frac{k}{n}\right)^{\frac{1}{n}}=\frac{1}{n}\sum_{k=1}^n\ln\frac{k}{n}
$$
tends to $-1$ as $n\to\infty$. Note that
$$
\int_1^n\frac{1}{n}\ln\frac{x}{n}\,dx\leq\sum_{k=1}^n\frac{1}{n}\ln\frac{k}{n}\leq\int_1^{n+1}\frac{1}{n}\ln\frac{x}{n}\,dx\,.
$$
Both integrals tend to $-1$ as $n\to\infty$, which completes the proof.
A: Since $(1+\frac{1}{n})^n\le e\le(1+\frac{1}{n})^{n+1}$, we have $(\frac{n+1}{e})^n\le n!\le(\frac{n+1}{e})^{n+1}$ by induction.
Hence, we can conclude that $\lim\limits_{n\to\infty}\frac{(n!)^{1/n}}{n}=\frac{1}{e}$.
A: Yet another way is to notice that the telescopic product
$$ \prod_{k=1}^{m-1}\left(1+\frac{1}{k}\right)=m $$
leads to
$$ n! = \prod_{m=2}^{n}\prod_{k=1}^{m-1}\left(1+\frac{1}{k}\right)=\prod_{k=1}^{n-1}\left(1+\frac{1}{k}\right)^{n-k}=\frac{n^n}{\prod_{k=1}^{n-1}\left(1+\frac{1}{k}\right)^k}, $$
such that
$$ \frac{n^n}{n!}=\prod_{k=1}^{n-1}\left(1+\frac{1}{k}\right)^k $$
and $\frac{n}{\sqrt[n]{n!}}$ is the geometric mean of $1,\left(1+\frac{1}{1}\right),\left(1+\frac{1}{2}\right)^2,\ldots,\left(1+\frac{1}{n-1}\right)^{n-1}$. This sequence is increasing and convergent to $e$, so $e$ has to be the wanted limit by Cesàro.
