$\lim\limits_{n \to{+}\infty}{\sqrt[n]{n!}}$ is infinite

$n! \geq (n/2)^{n/2}$ because half of the factors are at least $n/2$. Take $n$-th root.

By considering Taylor series, $\displaystyle e^x \geq \frac{x^n}{n!}$ for all $x\geq 0,$ and $n\in \mathbb{N}.$ In particular, for $x=n$ this yields $$ n! \geq \left( \frac{n}{e} \right)^n .$$

Thus $$\sqrt[n]{n!} \geq \frac{n}{e} \to \infty.$$

Using $\text{AM} \ge \text{GM}$

$$ \frac{1 + \frac{1}{2} + \dots + \frac{1}{n}}{n} \ge \sqrt[n]{\frac{1}{n!}}$$

$$\sqrt[n]{n!} \ge \frac{n}{H_n}$$

where $H_n = 1 + \frac{1}{2} + \dots + \frac{1}{n} \le \log n+1$

Thus $$\sqrt[n]{n!} \ge \frac{n}{\log n+1}$$