Proving $n! < n^n$ I have to prove $n! < n^n$ for positive integers greater than 1, but with a little twist.  I have to show $P(n-1)$ holds.  For the left, I know $(n-1)! = \frac{n!}{n}$ and I'm stuck from there thinking about $(n-1)^{n-1}$.
 A: The inductive proof is standard and will surely be supplied by another answer. Here's a cooler way to do it: 
Recall that $n! = \underbrace{n \cdot (n-1) \cdot (n-2) \cdot \ldots \cdot 2 \cdot 1}_{\text{n elements}}$.
Recall that $n^n = \underbrace{n \cdot n \cdot n \cdot \ldots \cdot n}_{\text{n elements}}$
So both $n!$ and $n^n$ are sequences of length $n$: for the first element of both, $n=n$. Thereafter, the element from $n^n$ is always larger, because $n > n-1, \ldots, n >1$. We know from the axioms of arithmetic and order for $\mathbb{Z}$ that if $a>b$ and $c>0$, then $ac > ab$. So the product of the elements of $n^n$ will be larger. 
A: HINT:
If  $ m^m> m!,$
$ (m+1)^{m+1}=(m+1) (m+1)^m> (m+1)m^m$ for $m>0$
$\implies (m+1)^{m+1} >(m+1)\cdot m!=(m+1)!$
A: Note that $$n^n=(n-1)^{n-1}\cdot \underbrace{\left(1+\frac1{n-1}\right)^{n-1}}_{>1\text{ if }n\ge 2}\cdot n$$
A: $1 \lt n$
$2 \lt n$
$3 \lt n$
...
$n-1 \lt n$
$n\le n$
If you multiply you get:
$1\cdot 2\cdot 3\cdot...\cdot (n-1)\cdot n \lt n\cdot n\cdot n\cdot...\cdot n\cdot n$
$n!\lt n^n$
A: Would it be acceptable to start from an obviously true statement like
$$n! < (n-1)^n + (n-1)!$$
and work backwards to your desired inequality.
$$n! < (n-1)^n + (n-1)!$$
$$n!-(n-1)! < (n-1)^n$$
$$\frac{n(n!)-n!}{n} < (n-1)^n$$
$$\frac{(n!)(n-1)}{n}<(n-1)^n$$
$$\frac{n!}{n}<\frac{(n-1)^n}{n-1}$$
$$(n-1)!<(n-1)^{n-1}$$
A: If you use induction, the inductive step is
$$(n+1)n!<(n+1)n^n<(n+1)(n+1)^n=(n+1)^{n+1}$$
