Find $\lim_{n \to \infty} \sqrt[n]{n!}$.

I'm not sure it holds in general. Consider $n! = n(n-1)(n-2)...2\cdot1$, at least $n/2$ terms are $>n/2$ thus; $$n! > \left(\frac{n}{2}\right)^\frac{n}{2}$$ taking the reciprocal $$ \frac{1}{n!} < \left(\frac{2}{n}\right)^\frac{n}{2}$$ then raising to the power $1/n$ gives $$ \left(\frac{1}{n!}\right)^\frac{1}{n} < \left(\frac{2}{n}\right)^\frac{1}{2}$$ clearly the right hand side has $\lim_{n\rightarrow \infty} \sqrt{2/n} = 0$ thus $$\lim_{n\rightarrow \infty} \left(\frac{1}{n!}\right)^\frac{1}{n} = 0$$

Edit: of course njguliyev's comment is a much easier way to see it can't be $\geq1$.

An alternative simple way to show that $\lim_{n\to \infty} \sqrt[n]{\frac1{n!}}=0$ may be as follows. Recall that for any sequence $\{c_n\}$ of positive numbers, we have $$\limsup_{n\to \infty} \sqrt[n]{c_n}\le \limsup_{n\to \infty} \frac{c_{n+1}}{c_n}.$$ Now with $c_n\triangleq \frac{1}{n!}$, clearly $\limsup_{n\to \infty} \frac{c_{n+1}}{c_n}=0$, which implies that $\limsup_{n\to \infty} \sqrt[n]{c_n}=0.$ Therefore $\lim_{n\to \infty} \sqrt[n]{c_n}=0,$ since it's obvious that $\liminf_{n\to \infty} \sqrt[n]{c_n}\ge 0.$

$$ \lim_{n \to \infty}{-\ln\left(\Gamma\left(n + 1\right)\right) \over n} = -\lim_{n \to \infty}\Psi\left(n + 1\right) = -\infty $$

$$ \color{#ff0000}{\lim_{n \to \infty}\left(1 \over n!\right)^{1/n} = 0} $$

$\Gamma$ and $\Psi$ are the Gamma and Digamma functions, respectively.

$\mbox{Or Stirling dixit:}$ \begin{align} \lim_{n \to \infty}\left(1 \over n!\right)^{1/n} & = \lim_{n \to \infty}\left(1 \over \sqrt{2\pi\,}\,n^{n + 1/2}\, {\rm e}^{-n}\right)^{1/n} = {1 \over \sqrt{2\pi\,}\,}\, \lim_{n \to \infty}{{\rm e} \over n^{1 + 1/2n}} \\[5mm] & = {{\rm e} \over \sqrt{2\pi\,}\,}\, \lim_{n \to \infty}\left(1 \over n\right) = \color{#ff0000}{0} \end{align}