How to prove $\left(\frac{n}{n+1}\right)^{n+1}<\sqrt[n+1]{(n+1)!}-\sqrt[n]{n!}<\left(\frac{n}{n+1}\right)^n$ 
Show that:
  $$\left(\dfrac{n}{n+1}\right)^{n+1}<\sqrt[n+1]{(n+1)!}-\sqrt[n]{n!}<\left(\dfrac{n}{n+1}\right)^n$$
  where $n\in \Bbb N^{+}.$

If this inequality can be proved, then we have
$$\lim_{n\to\infty}\sqrt[n+1]{(n+1)!}-\sqrt[n]{n!}=\dfrac{1}{e}.$$
But I can't prove this inequality. Thank you.
 A: Hint:
Applying Stolz–Cesàro theorem
$$L=\lim_{n\to \infty} \frac{\sqrt[n]{n!}}{n}=\lim_{n\to \infty} \frac{\sqrt[n+1]{(n+1)!}-\sqrt[n]{n!}}{(n+1)-n}=\lim_{n\to \infty} \left(\sqrt[n+1]{(n+1)!}-\sqrt[n]{n!}\right)$$
We have $$\ln L=\lim_{n\to \infty} \left(\frac{1}{n}\sum_{i=0}^{n}\ln\left(\frac{i}{n}\right)\right)=\int_{0}^{1}\ln x\,dx=-1$$
$$\to L=\frac{1}{e}$$
A: This does not answer the question completely, but proves the inequalities for large enough $n$. 
Let $f(x)=e^{g(x)}$ where $g(x)=\frac{1}{x}\log\Gamma(x+1)$, and $h(x)=\left(\frac{x}{x+1}\right)^x$.
It is enough to show that 
$$\frac{x}{x+1}h(x)<f'(x)<h(x),$$
since $f(n+1)-f(n)=f'(c_n)$ for some $n< c_n <n+1$ by the Mean Value Theorem.  
We use a version of Stirling's approximation of $\Gamma$ function:
$$\log\Gamma(x+1)=x\log x-x+\frac{1}{2}\log(2\pi x)+\sum_{n=1}^\infty \frac{B_{2n}}{2n(2n-1)x^{2n-1}}$$ 
This yields the following asymptotic relations for large enough $x$. 
$$g(x)=\log x-1+\frac{\log(2\pi x)}{2x}+O(\frac{1}{x^2}),$$
$$g'(x)=\frac{1}{x}+\frac{1}{2x^2}-\frac{\log(2\pi x)}{2x^2}+O(\frac{1}{x^3}),$$
$$f'(x)=e^{g(x)}g'(x)=\frac{1}{e}(1+\frac{1}{2x}-\frac{\log^2(2\pi x)}{8x^2}+O(\frac{\log x}{x^2})),$$
$$h(x)=\frac{1}{e}(1+\frac{1}{2x}+O(\frac{1}{x^2})),$$
$$\frac{x}{x+1}h(x)=\frac{1}{e}(1-\frac{1}{2x}+O(\frac{1}{x^2}))$$
Thus, we have the claim for large enough $x$, and this proves the inequalities for large enough $n$. 
Remark1) Treating error terms extra carefully might give an explicit $N$ such that the inequalities hold for $n>N$. 
Remark2) Once we find such $N$, we can check one by one for $n=1,2,\cdots N$. 
A: Here is a partial answer :
If a sequence $u=(u_n)_{n\ge1}$ of real numbers converges, then the sequence $\left(\frac{1}{n}\sum_{k=1}^nu_k\right)_{n\ge1}$ converges to the same limit. This is the well known Cesaro's lemma.
It can be proved that the converse is false (consider the sequence $u=((-1)^n)_{n\ge1}$) but becomes true if we assume that $u$ is monotonic.
As a consequence, we get the following result :
If $t=(t_n)_{n\ge1}$ is a monotonic sequence of real numbers such that $\lim_{n\to\infty}\frac{u_n}{n}=L\in\mathbb{R}$ then $\lim_{n\to\infty}\left(u_{n+1}-u_n\right)=L$.
Let's apply this last result to the sequence $t=(\left[n!\right])^{1/n})_{n\ge1}$.
It's easy to show ($\color{red}{\mathrm{see}\,\,\mathrm{below}}$) that
$$\lim_{n\to\infty}\frac{\left[n!\right]^{1/n}}{n}=\frac{1}{e}$$
Therefore, if we prove that $t$ is monotonic (at least ultimately, which is a sufficient condition), we will reach the conclusion that :
$$\lim_{n\to\infty}\left(\left[(n+1)!\right]^{1/(n+1)}-\left[n!\right]^{1/n}\right)=\frac{1}{e}$$

It is a consequence of Cesaro's lemma that if a sequence of positive real numbers $(x_n)_{n\ge1}$ verifies $\lim_{n\to\infty}\frac{x_{n+1}}{x_n}=L$ some $L>0$ then $\lim_{n\to\infty}\left(x_n\right)^{1/n}=L$.
Applying this to sequence $x_n=\left(\frac{n!}{n^n}\right)$ we obtain  $\lim_{n\to\infty}\frac{\left[n!\right]^{1/n}}{n}=\frac{1}{e}$
A: By partial summation:
$$\sum_{j=1}^{n}\log j = n\log n - n + \sum_{j=1}^{n}\left(1-j\log(1+1/j)\right),$$
so it is possible to fix @i707107's argument (removing "for any $N$ big enough") by noticing that
$$\left(1-j\log(1+1/j)\right) \in \left(\frac{1}{2j},\frac{1}{2j+2}\right).$$
