Proof $\frac{\log(n!)}{n\log n}$ is increasing for positive integer $n$ ($\log n=\ln(n)$)

Question

I would like to show that $\frac{\log(n!)}{n\log n}$ increases as n increases, for positive integer n only (to ignore the use of gamma function). Note here $\log n = \ln(n)$, so using base e. I would like to be able to show this so I can then apply the Monotone Convergence Theorem to show that $(\log(n!)\sim(n\log n)$. My first idea was to try and show that the second derivative is always positive, but due to $\log(n!)$ not being continuous (without having to use gamma function which I feel is OTT for this question) this, of course, would not work. Does anyone have any ideas about how to prove that this sequence is increasing rigorously?

Answer 1

Hint: You can use Stirling approximation for $n!$: $$n! \sim \sqrt{2\pi n}\left(\frac{n}{e}\right)^n$$ Or: $$n! = \sqrt{2\pi n}\left(\frac{n}{e}\right)^n \left(1+O\left(\frac{1}{n}\right)\right)$$ To know more about how to use this approximation you can see this link.