Find the limit $\lim _{n \to \infty}(n!)^{{\frac{1}{n^2}}}$

You can just squeeze it: $$ 1\leq(n!)^{1/n^2}\leq (n^n)^{1/n^2}=n^{1/n}\to 1. $$ So $\lim_{n\to\infty} (n!)^{1/n^2}=1$.

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

$L=e^\frac{\log_{e}{n!}}{n^2}=e^\frac{\sum_{i=1}^{n}{\log(i)}}{n^2}$

Now $0\leq\frac{\sum_{i=1}^{n}{\log(i)}}{n^2}\leq \frac{n\log(n)}{n^2}=\frac{\log(n)}{n}$ which tends to $0$ as $n$ tends to $\infty$. Hence$\lim\limits_{n \to \infty}{\frac{\sum_{i=1}^{n}{\log(i)}}{n^2}} = 0$.

$\lim\limits_{n \to \infty}{L} = e^{\lim\limits_{n \to \infty}{\frac{\sum_{i=1}^{n}{\log(i)}}{n^2}}}=e^0=1$

Let$$y=\lim _{n \to \infty}(n!)^{\large {\frac{1}{n^2}}}$$ Therefore $$\ln y=\lim _{n \to \infty}\frac{1}{n^2}\sum_{r=0}^{n-1}\ln(n-r)$$ $$\ln y=\lim _{n \to \infty}\frac{\ln n}{n^2} - \frac1{n}\sum_{r=0}^{n-1}\frac1n\ln(1-\frac{r}{n})$$ $$\ln y=0- \lim _{n \to \infty}\frac1{n}\int_{0}^{1}\ln(1-x)dx$$ $$\ln y=0- \lim _{n \to \infty}\frac1{n}(-1)$$ $$\ln y=0- 0$$ $$y=e^0=\boxed{1}$$