calculus - Show that $lim_{ntoinfty}frac{ln(n!)}{n} = +infty$

Show that
$$\lim_{n\to\infty}\frac{\ln(n!)}{n} = +\infty$$

The only way i've been able to show that is using Stirling's approximation:
$$n! \sim\sqrt{2\pi n}\left(\frac{n}{e}\right)^n$$

Let:
$$\begin{cases} x_n = \frac{\ln(n!)}{n}\\ n \in \Bbb N \end{cases}$$

So we may rewrite $x_n$ as:

$$x_n \sim \frac{\ln(2\pi n)}{2n} + \frac{n\ln(\frac{n}{e})}{n}$$

Now using the fact that $\lim(x_n + y_n) = \lim x_n + \lim y_n$ :
$$\lim_{n\to\infty}x_n = \lim_{n\to\infty}\frac{\ln(2\pi n)}{2n} + \lim_{n\to\infty}\frac{n\ln(\frac{n}{e})}{n} = 0 + \infty$$

I'm looking for another way to show this, since Stirling's approximation has not been introduced at the point where i took the exercise from yet.

Answer

Another way to show $$\lim_{n\to\infty}\frac{\ln(n!)}{n}=\infty$$ is to consider the following property of logarithms: $$\log(n!)=\log(n)+\log(n-1)+\cdots+\log(2)>\frac{n}{2}\log\left(\frac n2\right).$$ Now $$\frac{\log (n!)}{n}>\frac{\log(n/2)}{2}.$$ As $n\to\infty$, this clearly diverges to $+\infty$.