TITLE: Why $n!^{(1/n)}$ behaves like an arithmetic progression?
QUESTION [6 upvotes]: The sequence $(n!)^{(1/n)}$ is an increasing sequence. Here are some terms,
$$ 1,1.4142, 1.8171,2.2133, 2.6051,2.9937,....$$
Could someone explain why this sequence is so close to an arithmetic progression?

REPLY [3 votes]: The first step in an explanation is Stirling's formula:
$$n!\approx \sqrt{2\pi n}\left(\frac ne\right)^n$$
so that 
$$n!^{1/n}\approx (\sqrt{2\pi n})^{1/n}\frac ne.$$
The $(\sqrt{2\pi n})^{1/n}$ part rapidly approaches 1, so can be disregarded, and the common difference of your terms is simply $\frac1e\approx 0.367$.
This blog post of Terry Tao explains where Stirling's formula comes from.