Prove $\lim\limits_{n\to\infty}\frac{1}{\sqrt[n]{n!}}=0$ I used $$(n!)^{\frac{1}{n}}=e^{\frac{1}{n}\ln(n!)}=e^{\lim\limits_{n\to\infty}\frac{1}{n}\ln(n!)}$$
Then using Stirling's approximation and L'Hospital's rule on $$\lim\limits_{n\to\infty}\frac{\ln(n!)}{n}$$ I get $$\lim\limits_{n\to\infty}\frac{\ln(n!)}{n}=\lim\limits_{n\to\infty}(\ln(n)+\frac{n+\frac{1}{2}}{n}-1)=\infty$$
Now, $$e^{\lim\limits_{n\to\infty}\frac{1}{n}\ln(n!)}=e^{\infty}=\infty$$
Thus
$$\lim\limits_{n\to\infty}\frac{1}{\sqrt[n]n}=\frac{1}{\infty}=0$$
Is this correct approach and what other methods could be used?

As long as you haven't made an algebra mistake, stirlings approximation should work.
Separate $n!$ into two parts. Assign $1$ to everything below $n/2$. Assign $n/2$ to everything above $n/2$
So you get $(n/2)^{n/2}<n!$
Apply the root and you get:
$(n/2)^{1/2}<n!^{1/n}$
$(n/2)^{1/2}$ clearly aproaches infinity so it will make the limit zero.
