Answer
See proof
Work Step by Step
$$\lim\limits_{x \to \infty}\frac{e^x}{x^n}=\frac{\infty}{\infty}$$
Using the l'Hospital's rule it follows:
$$\lim\limits_{x \to \infty}\frac{e^x}{x^n}=\lim\limits_{x \to \infty}\frac{e^x}{nx^{n-1}}=\frac{\infty}{\infty}$$
Using the l'Hospital's rule again it follows:
$$\lim\limits_{x \to \infty}\frac{e^x}{x^n}=\lim\limits_{x \to \infty}\frac{e^x}{n(n-1)x^{n-2}}=\frac{\infty}{\infty}$$
The $n$-derivative of $e^{x}$ is $e^x$ and the $n$-derivative of $x^n$ is $n!$ so using the l'Hospital's rule $n$ times it follows:
$$\lim\limits_{x \to \infty}\frac{e^x}{x^n}=\lim\limits_{x \to \infty}\frac{e^x}{n!}=\frac{1}{n!}\lim\limits_{x \to \infty}e^x=\frac{1}{n!} \cdot \infty=\infty$$