Evaluate $\lim_{n \to \infty} (n!)^{\frac{1}{n^2}}$

Question

Let $a_n = (n!)^{\frac{1}{n^2}}$.

Now, $$n! \geq1 \implies (n!)^{\frac{1}{n^2}} \geq 1$$ and $$n! \leq n^n \implies (n!)^{\frac{1}{n^2}} \leq n^{\frac1n}$$

But $$\lim_{n \to \infty} n^{\frac1n} = 1 = \lim_{n \to \infty} 1$$ Thus by Sandwich Theorem $$\lim_{n \to \infty} (n!)^{\frac{1}{n^2}} =1$$ Is the solution correct?

Answer 1

Alternatively: $$ \lim_{n\to\infty}\log(n!)^{\frac{1}{n^2}} = \lim_{n\to\infty}\frac{\log(n!)}{n^2} = \lim_{n\to\infty}\frac{\log 1 + \log 2 + \cdots + \log(n)}{n^2} = $$ $$ \lim_{n\to\infty}\frac{\log(n + 1)}{(n + 1)^2 - n^2} = \lim_{n\to\infty}\frac{\log(n + 1)}{2n + 1} = 0. $$ (Stolz-Cesàro used in the 3rd $=$)