Is there a general way to find large enough $n \in \mathbb N$ such that for $k \in \mathbb N$: $n!>k^{n}$ 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.
