It is convincing that the limit is 0 since $n$ has one fixed increasing step 1 but $\log(n!)$ has one strictly increasing step $\log n$. This is by seeing the limit as $\lim_{n\to \infty}\frac{\sum_{k=1}^{n}1}{\sum_{k=1}^{n}\log k}$.
Wolfram Alpha tells me to use Stirling's approximation. Then is there one way to prove $\lim_{n\to \infty}\frac{n}{\log(n!)}=0$ without using Stirling's approximation? By the above step inspect, it seems that we can use Squeeze theorem where one direction $\frac{n}{\log(n!)}\ge 0$ is trivial.
Best Answer
We have that $$\dfrac{n}{\ln(n!)}=\dfrac{n}{\ln(2)+\ln(3)+\cdots+\ln(n)}$$
where both $a_n=n$ and $b_n=\ln(2)+\ln(3)+\cdots+\ln(n)$ are strictly increasing sequences that diverge to $+\infty$.
We can see that $$\dfrac{a_{n+1}-a_n}{b_{n+1}-b_n}=\dfrac{1}{\ln(n+1)}\to 0$$
By the Stolz-Cesàro theorem, $\dfrac{a_n}{b_n}\to 0$.