It is the strirling's aproximation See article
$n!\approx \left(\frac{n}{e}\right)^{n}\sqrt{2\pi n}$
The error estimates are very interesting:
$n!=\left(\frac{n}{e}\right)^{n}\sqrt{2\pi n} \cdot e^{r_{n}}$
With $\frac{1}{12n+1}
This is the strirling's aproximation with remainder
Edit: This aproximation is very useful, for example in the calculation of limits of sequences
For example,
$$\frac{\sqrt[n]{n!}}{n} \approx \frac{\sqrt[n]{\left(\frac{n}{e}\right)^{n}\sqrt{2\pi n}}}{n} = \frac{\left(\frac{n}{e}\right)\sqrt[n]{\sqrt{2\pi n}}}{n} = \frac{\sqrt[2n]{2\pi n}}{e} \longrightarrow \frac{1}{e}, ( n\longrightarrow \infty ) $$
Using the fact $\sqrt[n]{n} \longrightarrow 1, ( n\longrightarrow \infty )$