Im trying to prove that $\forall{m}\in \mathbb{N} - \{ 1 \}. \exists B_m \in \mathbb{N}. \forall{n} \in \mathbb{N}. n \geq B_m \implies n! > m^{B_m}$
There seems to be a pattern. For $n=2 \rightarrow B_2=4$, $n=3 \rightarrow B_3=7$, $n=4 \rightarrow B_4=9$, $n=5 \rightarrow B_5=12$ and $n=6 \rightarrow B_6=14$
The pattern I'm seeing is that for all odd and even numbers we have that $B_2=4$ and $B_3=9$ then for all even/odd numbers we have that $B_m = B_{m-2} + 5$. That has been my approach for know. Simply try to prove this by induction on even numbers first and then odd and conclude that the statement is proven. However, I'm struggling quite a bit with the main induction proof on $m$.
The first induction proof is quite simple. We have to prove that given an $B_m \in \mathbb{N}. \forall n \in \mathbb{N}. n \geq B_m \implies n! > m^n$. This is fairly simple. Since $B_m>m$ it must be that $(n+1)\times n!> m^{n+1} $ holds.
What remains to be proven then is $B_m! > m^{B_m}$ by induction on $m$. I'm struggeling a lot on this step and was hoping someone could give me a nudge in the right direction. My approach of splitting the set into even and odds might be totally wrong so feel free to point out any flaws in my current logic!