# How can I prove this statement by mathematical induction?

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.

• You're on the right track: $n!(n+1) \leqslant n^n(n+1) \leqslant(n+1)^n(n+1)=(n+1)^{n+1}$
– M.P
Jun 19, 2019 at 9:54
• A simple search can get you many relevant answers for common questions. Jun 19, 2019 at 9:55
• Here is a video explaining: youtu.be/NsO6nh42oPo Jun 19, 2019 at 9:56
• Another approach: Note that $\frac{n!}{n^n} = \prod_{k=1}^n \frac{k}{n}$. Then, use induction to prove that the product of $n$ numbers less than $1$ is also less than $1$. Jun 19, 2019 at 9:59

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^{n}<(n+1)^{n}$$ So $$(n+1)(n^{n})<(n+1)^{n+1}$$. Hence, if we assume that $$n! we get $$(n+1)!=(n+1) n! <(n+1) n^{n}<(n+1)^{n+1}$$