This completes Jonas's answer, here is an idea. This is too long for a comment.
To prove that the limit exists, we can prove that $a_n$ is decreasing and positive:
$\frac{1}{\sqrt[n+1]{(n+1)!} - \sqrt[n]{(n)!}} \geq 0 \Leftrightarrow $
$\sqrt[n+1]{(n+1)!} \geq \sqrt[n]{(n)!} \Leftrightarrow $
$(n+1)!^n\geq (n)!^{n+1} \Leftrightarrow $ $(n+1)^n\geq (n)! \Leftrightarrow $ $(n+1)\cdot(n+1)...\cdot(n+1)\geq 1\cdot 2..\cot n \checkmark $
Now for decreasing
$\frac{1}{\sqrt[n+1]{(n+1)!} - \sqrt[n]{(n)!}} \geq \frac{1}{\sqrt[n+2]{(n+2)!} - \sqrt[n+1]{(n+1)!}} \Leftrightarrow $
$\sqrt[n+1]{(n+1)!} - \sqrt[n]{(n)!} \leq \sqrt[n+2]{(n+2)!} - \sqrt[n+1]{(n+1)!} \Leftrightarrow $
$2\sqrt[n+1]{(n+1)!} \leq \sqrt[n]{(n)!}+ \sqrt[n+2]{(n+2)!}$
Now, by AM-GM inequality
$\frac{\sqrt[n]{(n)!}+ \sqrt[n+2]{(n+2)!}}{2} \geq (n!)^\frac{1}{2n}[(n+2)!]^\frac{1}{2n+1}$
So if we can prove that
$(n!)^\frac{1}{2n}[(n+2)!]^\frac{1}{2n+4} \geq \sqrt[n+1]{(n+1)!}$
we are done.
Now
$(n!)^\frac{1}{2n}[(n+2)!]^\frac{1}{2n+4} \geq \sqrt[n+1]{(n+1)!} \Leftrightarrow $ $(n!)^\frac{1}{2n}[(n+2)]^\frac{1}{2n+4} \geq (n+1)!^{\frac{1}{n+1}-\frac{1}{2n+4}} \Leftrightarrow $
$(n!)^{\frac{1}{2n}+\frac{1}{2n+4}-\frac{1}{n+1}}[(n+2)]^\frac{1}{2n+4} \geq (n+1)^{\frac{1}{n+1}-\frac{1}{2n+4}}\,. $
To keep it simple:
The power of $n!$ is
$\frac{(2n^2+6n+4)+(2n^2+2n)-(4n^2+8n)}{(n+1)2n(2n+4)}=\frac{2}{n(n+1)(2n+4)}$
The power of $n+1$ is
$\frac{1}{n+1}-\frac{1}{2n+4}=\frac{n+3}{(n+1)(2n+4)}$
Thus, after bringing the inequality to the $n(n+1)(2n+4)$, it becomes:
$(n!)^\frac{1}{2n}[(n+2)!]^\frac{1}{2n+4} \geq \sqrt[n+1]{(n+1)!} \Leftrightarrow $ $(n!)^2(n+2)^{n(n+1)} \geq (n+1)^{n(n+3)} $
Now, I ma not sure that this is true, but might work....