I have been trying to prove that $e^{x} \geq x^{e}$ for all $x \geq 0$.
This is what I have gotten so far:
$$ f(x) = e^{x} \\ g(x) = x^{e} \\ f'(x) = \frac{d(e^{x})}{dx} = e^{x} \\ g'(x) = \frac{d(x^{e})}{dx} = e x^{e-1}\\ $$ Now for all $x \geq 0$, $f'(x) \geq 0$, so $f(x)$ is always increasing or is constant, and $g'(x) \geq 0$ so $g(x)$ is always increasing or is constant.
Thus across the interval $[0,e]$, the lowest $f(x)$ can be is $f(0) = 1$ and the lowest $g(x)$ can be is $g(0) = 0$. The highest possible value of $f(x)$ is $f(e) = e^{e}$ and highest value of $g(x)$ is $g(e) = e^{e}$.
Thus across the interval $[0,e]$, $f(x) \geq g(x)$.
I've got 2 questions. One, is what I did correct. If not, where did I go wrong and what is the right method. Two, if yes, how does one prove it for the interval $(e,\infty)$.
Your proof is incorrect. Just because both functions are increasing doesn't mean that the one that is ahead always maintains the lead. This is because $g(x)$ could pass $f(x)$ because it starts off increasing sufficiently fast, and then at a later point $f(x)$ could start increasing very rapidly to regain the lead over $g(x)$.
EDIT: a specific counter example has been given in the comments.