I'm having trouble proving that $$n! \leqslant n^n \, \, \, \,\forall \,n \in \mathbb{Z}^+$$ by mathematical induction. I checked if it worked for $n = 1$ and then supposed that it worked for $n$, to then prove if it worked for $n+1$.
In this last step I tried writing $(n+1)!$ like $n!(n+1)$ but I don't know how to continue. Thank you so much.
Let $n! \le n^n$.
Since $n^n \le (n+1)^n$ we get
$$(n+1)!=n!(n+1) \le n^n(n+1) \le (n+1)^n(n+1) = (n+1)^{n+1}.$$
$n! (n+1) \le n^n(n+1)$ by induction assumption, and $n^n(n+1) < (n+1)^n(n+1)$ because $n<n+1$.
$n^{n}<(n+1)^{n}$ So $(n+1)(n^{n})<(n+1)^{n+1}$. Hence, if we assume that $n! <n^{n}$ we get $(n+1)!=(n+1) n! <(n+1) n^{n}<(n+1)^{n+1}$
$n!(n+1) \leqslant n^n(n+1) \leqslant(n+1)^n(n+1)=(n+1)^{n+1}$ – M.P Jun 19 '19 at 09:54