How to solve this question related to limits of successions? I've done these two proves: 
$\left(\frac{\sqrt[n]{n!}}{n}\right)_{n}\rightarrow \frac{1}{e}$
$\left(\frac{\sqrt[n+1]{(n+1)!}}{\sqrt[n]{n!}}-1 \right)_{n} \rightarrow 0$
And now I've to prove this statement, using the previous statements if are necessary:
$\left(\frac{\sqrt[n+1]{(n+1)!}}{\sqrt[n]{n!}} \right)_{n}^{n}\rightarrow e$
All I could done to solve the limit is transformate the expression in this way:
$e^{lim_{n \to \infty} n\left(\frac{\sqrt[n+1]{(n+1)!}}{\sqrt[n]{n!}}\right)}$
and theoretically the limit of the exponent has to be 1, but I dont't know how to continue to prove it.
Thanks in advance.
 A: Alternatively, by (a weak) Stirling approximation, as $n \to \infty$,
$$
\ln n!=n\ln n-n+O(\ln n)
$$ one may write, as $n \to \infty$,
$$
\begin{align}
\left(\frac{\sqrt[n+1]{(n+1)!}}{\sqrt[n]{n!}} \right)^{n}&=\left(\frac{\sqrt[n]{n+1}\times\sqrt[n+1]{(n+1)!}}{\sqrt[n]{(n+1)!}} \right)^{n}
\\\\&=(n+1)\times ((n+1)!)^{-\frac1{n+1}}
\\\\&=e^{\ln(n+1)}\times e^{\large-\frac{\ln[(n+1)!]}{n+1}}
\\\\&=e^{\ln(n+1)}\times e^{\large-\ln(n+1)+1+O\big(\frac{\ln n}n\big)}
\\\\&=e\times e^{O\big(\frac{\ln n}n\big)}
\\\\& \to e
\end{align}
$$
as expected.
A: It is sufficient to show that
$$ \lim_{n\to\infty}\left(\frac{\sqrt[n+1]{(n+1)!}}{\sqrt[n]{n!}}-\frac{n+1}{n}\right)=0.$$
Since $\lim_{n\to\infty}\frac{\sqrt[n]{n!}}{n}=\frac1e$, one has
\begin{eqnarray}
\lim_{n\to\infty}\left(\frac{\sqrt[n+1]{(n+1)!}}{\sqrt[n]{n!}}-\frac{n+1}{n}\right)&=&\lim_{n\to\infty}\left(\frac{(n+1)\frac{\sqrt[n+1]{(n+1)!}}{n+1}}{n\frac{\sqrt[n]{n!}}{n}}-\frac{n+1}{n}\right)\\
&=&\lim_{n\to\infty}\left(\frac{n+1}{n}\right)\left(\frac{\frac{\sqrt[n+1]{(n+1)!}}{n+1}}{\frac{\sqrt[n]{n!}}{n}}-1\right)\\
&=&0
\end{eqnarray}
and hence
$$　\frac{\sqrt[n+1]{(n+1)!}}{\sqrt[n]{n!}}＝1+\frac{1}{n}＋o(\frac1{n}).　$$
So
$$　\left(\frac{\sqrt[n+1]{(n+1)!}}{\sqrt[n]{n!}}\right)^n＝\left(1+\frac1{n}+o(\frac1{n})\right)^n.　$$
Therefore
$$　\lim_{n\to\infty}\left(\frac{\sqrt[n+1]{(n+1)!}}{\sqrt[n]{n!}}\right)^n＝e.　$$
A: $\left(\frac{\sqrt[n+1]{(n+1)!}}{\sqrt[n]{n!}} \right)^{n}=\frac{\sqrt[n+1]{(n+1)!^{n}}}{n!}=\sqrt[n+1]{\frac{(n+1)!^{n}}{n!^{n+1}}}=\sqrt[n+1]{\frac{(n+1)^{n}}{n!}}$
We know that if $lim_{n \to \infty} \frac{a_{n}}{a_{n-1}} $ exists, then $lim_{n \to \infty} \sqrt[n]{a_{n}}=lim_{n \to \infty} \frac{a_{n}}{a_{n-1}} $
So, applying this to $a_{n}:=\frac{(n+1)^{n}}{n!}$ we obtained:
$\frac{(n+1)^{n}}{n!} \frac{(n-1)!}{n^{n-1}}=\frac{n+1}{n}\left(\frac{n+1}{n}\right)^{n-1}=\left( 1+\frac{1}{n}\right)^{n} \rightarrow e$
