real analysis - Prove that $lim_{uto infty}{frac{u^m}{e^u}}=0$

I'm trying to prove that $\lim_{u\to \infty}{\frac{u^m}{e^u}}=0$ for any integer $m$. I tried using the Ratio Test For Limits to prove that the limit converges to $0$ for all real values of $m$, and therefore it must converge to $0$ for all integer values of $m$ since $\mathbb{Z} \subset \mathbb{R}$. However, using the Ratio Test only seems to complicate things. Is there a better way to show this?

Additionally, I need to prove that $\lim_{x\to 0^+}{x\log x}=0$ by setting $u=-\log x$ and using the above limit. So far, I have done the following:

$$\lim_{-\log x\to \infty}{\frac{(-\log x)^m}{e^{-\log x}}}=\lim_{x\to 0^+}x{(-\log x)^m}$$

but I don't know where to go from here. Also, I cannot use L'Hopital's rule, since I haven't covered it in lectures. Thanks in advance.

Answer

If you accept that $e^{u}=\sum_{n=0}^{\infty} \frac{u^n}{n!}$, then

$$\frac{u^m}{e^u}=\frac{u^m}{\sum_{n=0}^{\infty} \frac{u^n}{n!}}\le (m+1)!\frac{u^m}{u^{m+1}}\to 0$$

as $u\to\infty$.