I have found $n \in \mathbb N$ large enough such that $n!>2^{n}$ and $n!>5^{n}$ by simply "plugging in values". These facts have led me to consider a more general solution of $n!>k^{n}$ where $k \in \mathbb N$. In my attempts I have considered a function $f_{k}(n):=n!-k^{n}$ but it is discrete which makes working with derivatives painful. Is there a general solution and if so, how do I solve it?
I also thought of the following:
$n! > k^{n}\iff \log(n!)>n\log(k)\iff \frac{\log(n!)}{n}>\log(k)$
let's consider $$\sum_{n\in\mathbb{N}}\frac{k^n}{n!}$$ for some $k\in\mathbb{R}$ the sum is convergent Ratio test. Therefore $$\frac{k^n}{n!}\:\xrightarrow[]{n\to\infty}\:0$$ Thus, from the definition of the limit assuming $M=1$, there exists such $N$ for which when $n>N$ takes place $\frac{k^n}{n!}<1$ so $k^n<n!$
(edit after the commentary): Using Stirling's approximation we can write:
$$n!= \sqrt{2 \pi n} \left( \frac{n}{e} \right)^ne^{ \alpha_n} \ge \left( \frac{n}{e} \right)^n$$ the right side is greater than $k^n$ when $n/e>k$ so $n>ek$. This is a good estimate of practically the smallest $N_k=ek$ for which the above is the case.