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

Here is another way considering 


*

*$e^{\frac{\ln n!}{n}}$
\begin{eqnarray*} e^{\frac{\ln n!}{n}}
  & = &  \sqrt[n]{n!}\\
  &  \stackrel{GM-HM}{\geq} & \frac{n}{\frac{1}{1}+\cdots \frac{1}{n}} \\
  &   \stackrel{\sum_{k=1}^n \frac{1}{k} < \ln n + 1}{>}  & \frac{n}{\ln n +1} \\
 & \stackrel{n \to \infty}{\longrightarrow} & +\infty
\end{eqnarray*}
