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:
$n! > k^{n}\iff \log(n!)>n\log(k)\iff \frac{\log(n!)}{n}>\log(k)$
Since $n! > (n/e)^n$, if $n/e > k$ then $n!> k^n$.
Therefore $n > ek$ will work.
Consider the equation $$n! = k^n$$
If you look at this question of mine, you will see a magnificent approximation which was proposed by @robjohn. Adapted to your problem, it gives
$$\color{blue}{n\sim e\,k\,\exp\left(W\left(-\frac{\log (2 \pi k)}{2 e k}\right)\right)-\frac 12}$$ where $W(.)$ is Lambert function. For sure, you will need to use $\lceil n \rceil$.
For illustration purposes, let us try for $k=2^p$ and admire the results $$\left( \begin{array}{ccc} p & \text{approximation} & \text{exact} \\ 1 & 3.44447349465 & 3.45986564404 \\ 2 & 8.60959968277 & 8.61514776083 \\ 3 & 19.1868478144 & 19.1891973308 \\ 4 & 40.6214371321 & 40.6225104871 \\ 5 & 83.7910580611 & 83.7915684276 \\ 6 & 170.445206724 & 170.445454817 \\ 7 & 344.078832367 & 344.078954473 \\ 8 & 691.678759482 & 691.678820001 \\ 9 & 1387.21631441 & 1387.21634452 \\ 10 & 2778.63245721 & 2778.63247222 \\ 11 & 5561.80792391 & 5561.80793141 \\ 12 & 11128.5033891 & 11128.5033929 \\ 13 & 22262.2396826 & 22262.2396845 \\ 14 & 44530.0581347 & 44530.0581356 \\ 15 & 89066.0412025 & 89066.0412025 \end{array} \right)$$
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.
