I am able to establish that $\lim_{n\to \infty}\frac{1}{ n! ^\frac{1}{n} } $ converges to 0.
I suspect that $\frac{1}{ (n!)!^\frac{1}{n!} } $ is a subsequence of $ \frac{1}{ n! ^\frac{1}{n} } $ which would make the problem much easier to deal with. Is this a valid approach to the question?
We have that
$$\frac{1}{ (n!)!^\frac{1}{n!} }=e^{\frac{\log (n!)!}{n!}}$$
and then by $m=n!$ and by $m!\ge \frac{m^m}{e^m}$
$$\frac{\log (n!)!}{n!}=\frac{\log m!}{m}\ge \log \frac{m}{e}$$
Hint: write down$$\lim_{n\to \infty}\frac{1}{ (n!)!^\frac{1}{n!} }{=\lim_{m\to \infty}\frac{1}{ (m)!^\frac{1}{m} } } $$and apply Stirling's approx. for $m!$.
Yes, this is a valid approach to the question, because $(n!)_{n\in\Bbb N}$ is a subsequence of $(n)_{n\in\Bbb N}$. And if a sequence converges, then all of its subsequences converge too.
