show that sequence $a_n ={{n+1}\over \sqrt[n]{n!}}$ converges to e. if $e=\lim_{n \to \infty}{(1+1/n)^n} $ and  $a_n ={{n+1}\over \sqrt[n]{n!}}$ show that $
\lim_{n \to \infty}{a_n}= e $.
My teacher say that, I should use the next lemma
if ${a_n}$ converges to $a>0$ and for all i $a_i>0$ then 
$\lim_{n \to \infty}{\sqrt[n]{a_1a_2...a_n}}=a$
Thanks for you help.
 A: If you accept a solution that uses the natural logarithm then we may write,
$$\ln a_n=\ln (n+1)-\frac{1}{n} \ln n!$$
$$=\ln (n+1)-\frac{1}{n} \sum_{k=1}^{n} \ln k$$
$$=\ln (n+1)-\ln n-\frac{1}{n} \sum _{k=1}^{n} \ln (\frac{k}{n})$$
$$=\ln (1+\frac{1}{n})-\frac{1}{n} \sum _{k=1}^{n} \ln (\frac{k}{n})$$
As $n \to \infty$ the fist term vanishes and we have,
$$\to -\int_{0}^{1} \ln x dx$$
$$=1$$
So $a_n \to e$.
A: Observe that $a_0=1$.
For $n\ge 0$, let $b_n = a_n^n = \frac{(n+1)^n}{n!}$. Now $$b_{n}/b_{n-1} = \frac{(n+1)^{n}}{n^n}=(1+\frac{1}{n})^n\to e.$$
Finally, 
$$a_n^n = b_n = \prod_{j=1}^{n} \frac{b_j}{b_{j-1}}\quad \Rightarrow \quad a_n = \sqrt[n] {\prod_{j=1}^{n}\frac{b_j}{b_{j-1}}}\to e.$$
A: This is pretty much the proof of Stirling's formula in disguise, but:
$$
\frac1n\log(n!) - \log n = \bigg( \frac1n \sum_{j=1}^n \log j \bigg) - \log n = \frac1n \sum_{j=1}^n (\log j - \log n) = \frac1n \sum_{j=1}^n \log \frac jn,
$$
which is a Riemann sum:
$$
\lim_{n\to\infty} \bigg( \frac1n\log(n!) - \log n \bigg) = \lim_{n\to\infty} \frac1n \sum_{j=1}^n \log \frac jn = \int_0^1 \log x\,dx = (x\log x-x)\big|_0^1 = -1.
$$
(You can check that $\lim_{x\to0+} x\log x=0$.) Therefore
\begin{align*}
\lim_{n\to\infty} \frac{n+1}{\sqrt[n]{n!}} &= \lim_{n\to\infty} \frac{n+1}n \lim_{n\to\infty} \exp\bigg( \log n - \frac1n\log(n!) \bigg) \\
&= 1\cdot \exp\bigg( {-} \lim_{n\to\infty} \bigg( \frac1n\log(n!) - \log n \bigg) \bigg) = \exp(-(-1)) = e.
\end{align*}
