Showing that $\lim_{x \rightarrow \infty}\frac{e^x}{x^n} = \infty$ I have been told that I can show this by showing two things, first that
$$f(x) = \frac{e^x}{x^n}>\frac{e^n}{n^n}, \quad (x > n)$$
then
$$f'(x) = \frac{e^x(x-n)}{x^{n+1}}>\frac{e^{n+1}}{n^{n+1}}, \quad (x > n+1)$$
I have managed to show both of these things, the first is fairly easy and the second follows
$$\frac{e^x(x-n)}{x^{n+1}} = \frac{e^x}{x^n} - \frac{n}{x}\frac{e^x}{x^{n}} > \frac{e^x}{x^n}\bigg(\frac{1}{n+1}\bigg) > \frac{e^{n+1}}{n^{n+1}}$$
but I do not see how this helps me deduce that 
$\lim_{x \rightarrow \infty}\frac{e^x}{x^n} = \infty$
Just to clarify, I know how to show this a few other ways, I am interested in showing this way!

$e^x =$
$1 + x + \frac{x^2}{2!} + ...\frac{x^{n+1}}{(n+1)!} .....$
Hence: 
$e^x \gt \frac{x^{n+1}}{(n+1)!}$ , for $x \gt 0$, $ n \in \mathbb{N}$, $n \ge 1$.
$\frac{e^x}{x^n} \gt \frac{x}{(n+1)!}$.
Finally:
$\lim_{x \rightarrow \infty} \frac{e^x}{x^n} \ge \lim_{x \rightarrow \infty} \frac{x}{(n+1)!} = \infty$.
