TITLE: Find the limit $\lim _{n \to \infty}(n!)^{{\frac{1}{n^2}}}$
QUESTION [7 upvotes]: Find the limit $$\lim _{n \to \infty}(n!)^{{\frac{1}{n^2}}}$$
I have tried it using Stirling's approximation and then using L'Hospital's Rule and get answer $1$
Is there any other easy method to find this limit ?
Any hint. Thanks in advance.

REPLY [4 votes]: $$(n!)^{\frac{1}{n^2}} \le (n^n)^{\frac{1}{n^2}}=n^{\frac{1}{n}}=e^{\frac{\ln n}{n}}$$
Since
$$1 \le (n!)^{\frac{1}{n^2}} \le e^{\frac{\ln n}{n}}$$ and $\lim \limits_{n \to \infty} e^{\frac{\ln n}{n}}=1$ then by squeeze theorem there is a limit $\lim \limits_{n \to \infty} (n!)^{\frac{1}{n^2}}=1$