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.

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  $\lim_{n\to\infty}\frac{\left[n!\right]^{1/n}}{n}=\frac{1}{e}$
