If so, how can it be proven? (I have evaluated it up to $n=25$.)
If not, does there exist a $k\in\mathbb{R}$ such as that $n> a^k\Rightarrow n!>a^n$, with $n\in\mathbb{N},a\in\mathbb{R}$?
It is true, and to prove it, it suffices to show that $n!^2\geq n^n$ with induction.
For $n=1$ we have that $1\geq1$, which is true. Suppose $n!^2\geq n^n$. Then, $(n+1)!^2=(n+1)^2n!^2\geq (n+1)^2n^n$. We need to show that
$\begin{align} &(n+1)^2n^n\geq (n+1)^{n+1}\Leftrightarrow\\ \Leftrightarrow &n^n \geq (n+1)^{n-1}\Leftrightarrow\\ \Leftrightarrow &\ln(n^n)\geq \ln((n+1)^{n-1})\Leftrightarrow\\ \Leftrightarrow &n\ln n\geq (n-1)\ln(n+1)\Leftrightarrow\\ \Leftrightarrow &\frac{n}{n-1} \geq \ln(n+1-n)=\ln1=0\quad, \end{align}$
which is true!!