Correct usage of Stolz-Cesaro theorem in finding a limit Is this solution correct? 

I've seen this solution before, but I don't know why it is so. And which rule? 

The task is to find: $\omega=\displaystyle\lim_{n\to +\infty}\left(\sqrt[n+1]{(n+1)!}-\sqrt[n]{n!}\right)$

I know that this $\lim$ equals $\frac{1}{e},$ but do you see if this solution is correct or not? 

Solution is : 

$\displaystyle\lim_{n\to +\infty}\left(\sqrt[n+1]{(n+1)!}-\sqrt[n]{n!}\right)=\displaystyle\lim_{n\to +\infty}\left((n+1)\sqrt[n+1]{\frac{(n+1)!}{(n+1)^{n+1}}}-n\sqrt[n]{\frac{n!}{n^{n}}}\right)$

$=\displaystyle\lim_{n\to +\infty}\frac{\frac{(n+1)!}{(n+1)^{n+1}}}{\frac{n!}{n^{n}}}$ 

$=\frac{1}{e}$

Notes &amp; suggestions of mine:

I think he uses Stolz-Cesaro theorem: 

See that 

$\displaystyle\lim_{n\to +\infty}\frac{a_{n}}{b_{n}}=\displaystyle\lim_{n\to +\infty}\frac{a_{n+1}-a_{n}}{b_{n+1}-b_{n}}$ 

Chosen $a_{n}=n\sqrt[n]{\frac{n!}{n^n}}$ 

and $b_{n}=n$  clearly $b_{n}$ goes to $+\infty$ and $a_{n}$ goes to $+\infty$ because $n.\frac{1}{e}=+\infty$

we obtain: 

$\omega=\displaystyle\lim_{n\to +\infty}\frac{n\sqrt[n]{n!}}{n}$ 

Then use Cauchy-d'Alembert 

$=\displaystyle\lim_{n\to +\infty}\frac{\frac{(n+1)!}{(n+1)^{n+1}}}{\frac{n!}{n^{n}}}=\frac{1}{e}$ 

Is my explanation O.K.? 
May I ask for a correction?

Any remark ?
The use of Cesaro-Stolz as you mention is invalid. In application of Cesaro-Stolz the existence of limit of $\dfrac{a_{n+1}-a_n}{b_{n+1}-b_n}$ is a hypotheses and not the conclusion. Thus you have tried to use Cesaro-Stolz in reverse which does not work.
I don't think Stolz-Bolzano theorem is necessary here (:  , but I agree with everything you explained. Nevertheless, I'll write down everything in detail for everyone who might have not understood similar proofs to other (similar) posts.

Since $n\to\infty$, $n+1$ behaves (more or less) similarly:

Inductively: $$\lim_{n\to\infty}{\sqrt[n+1]{\frac{(n+1)!}{(n+1)^{n+1}}}}=\lim_{n\to\infty}{\sqrt[n]{\frac{n!}{n^n}}}$$
$$\implies\lim_{n\to +\infty}\left(\sqrt[n+1]{(n+1)!}-\sqrt[n]{n!}\right)=\lim_{n\to +\infty}\left((n+1)\sqrt[n+1]{\frac{(n+1)!}{(n+1)^{n+1}}}-n\sqrt[n]{\frac{n!}{n^{n}}}\right)=\lim_{n\to\infty}\Bigg((n+1)\sqrt[n]{\frac{n!}{n^n}}-n\sqrt[n]{\frac{n!}{n^n}}\Bigg)=\lim_{n\to\infty}{\sqrt[n]{\frac{n!}{n^n}}}=\lim_{n\to\infty}\frac{n!}{n^n}$$ 
As proven here:
Why is $\lim\limits_{n \to +\infty }{\sqrt[n]{a_1 a_2 \ldots a_n}} =\lim\limits_{n \to +\infty}{a_n}$

You can apply the Cauchy-D'Alembert criterion already here:$$\lim_{n\to\infty}\frac{\frac{(n+1)!}{(n+1)^{n+1}}}{\frac{n!}{n^n}}=\lim_{n\to\infty}\frac{n^n(n+1)!}{(n+1)^{n+1}n!}=\lim_{n\to\infty}\frac{n^n(n+1)}{(n+1)^{n+1}}=\lim_{n\to\infty}\frac{n^n}{(n+1)^n}=\lim_{x\to\infty}\frac{1}{\Big(\frac{n+1}{n}\Big)^n}=\lim_{n\to\infty}\frac{1}{\Big(1+\frac{1}{n}\Big)^n}=\frac{1}{e}$$