Inductive Proof of $n! < n^n$ Looking at the Wikipedia page on Mathematical Induction, I see that $n! < \frac{n^n}{2^n}\; \forall n>6$
I have been trying to prove that $n! < n^n \; \forall n>5$ using induction myself and my result is a clear special case of the above inequality. Does anyone have any hints on how to show either of these using induction? This came up when a student inquired about proving the convergence of a series, and induction seemed easier to show than a proof using Stirling's Approximation, but I can't seem to find a proof.
 A: Hint: Show $(n+1)n^n < (n+1)^{n+1}$.
A: Actually, $n!<n^n$ for $n>1$. The case $n=2$ is proved by inspection.
Suppose $n!<n^n$, for $n\ge2$; then $(n+1)!=(n+1)n!<(n+1)n^n$.
Moreover, from $n<n+1$, we deduce $n^n<(n+1)^n$. So we're done.

For $n!<n^n/2^n$, it's true only for $n>5$: indeed $6!=720$, while $6^6/2^6=729$.
So assume $n!<n^n/2^n$, for $n\ge6$. Then
$$
(n+1)!=(n+1)n!<(n+1)\frac{n^n}{2^n}
$$
and so we need to prove that
$$
(n+1)\frac{n^n}{2^n}\le\frac{(n+1)^{n+1}}{2^{n+1}}
$$
which is equivalent to
$$
2n^n\le (n+1)^n=n^n+\binom{n}{1}n^{n-1}+\sum_{k=2}^n\binom{n}{k}n^{n-k}
$$
and this is clearly true.
A: Fill in the details:
$$(n+1)!=(n+1)n!<(n+1)n^n<(n+1)(n+1)^n=(n+1)^{n+1}$$
A: $$\underbrace{n!=1*2*3\; ...*\;n}_{n\text{ digits}}$$
$$\underbrace{n^n=n*n*n\;...*\;n}_{n\text{ digits}}$$
Start with the first digit in each and proceed until reaching $n$.
$$1<n$$
$$2<n$$
$$3<n$$
$$\vdots$$
$$n=n$$
This should make it obvious that $n!<n^n$.
A: $(1+1+...+1)^{n}$ use multinomial expansion formula
Then $(1+1+...+1)^{n}>{n\choose{1}{1}{...}{1}}$
