To prove $a^x> x^a$ for all $x>c$, for some real $c$. ($a>1$).
Proof Attempt:
We first try to prove that $\lim_{x \to \infty}\frac{a^x}{x^a}=\infty$. Both of the functions in the numerator and the denominator goes to $\infty$ as $x\to\infty$.
So, L'Hospital's rule is applicable. We get $\lim_{x \to \infty}\frac{a^x \log (a)}{ax^{a-1}}$.
We repeat the operation $\lceil a \rceil = p$ times. We get $\lim_{x \to \infty}\frac{a^x(\ln (a))^{p} x^{p-a}}{a(a-1)(a-2)...(a-(p-1))} =\infty$
A function $f(x)$ is said to diverge to positive infinity, if $\forall G>0$, $\exists k>0$ such that $f(x)>G$, $\forall x>k$. We fix $G=1$ here. By the very definition, we can find a $k>0$ , such that
$\frac{a^x}{x^a}>1$ , $\forall x>k$, i.e. $a^x> x^a, \forall x>k$.
Is this method valid?
It seems valid, but you can make it a bit easier:
$$a^x>x^a \iff x>log_a(x^a)\iff x>alog_a(x)$$
I believe this case is much simpler to prove.