Show that $n^n>(n+1)!$ for all $n\ge3$ Show that $n^n>(n+1)!$ for all $n\ge3$
For $n=3$ it is to prove. 
assumed it true for some fixed $n\in \mathbb{N}$.
Then tried to prove for $n+1$
$(n+1)^{n+1}=(n+1)^n(n+1)>n^n(n+1)>(n+1)!(n+1)$
Got stuck.
 A: $$\frac{n^n}{n!}=\frac n1\cdot\frac n2\cdot\frac n3\cdots\frac nn>\frac n1\cdot\frac n2>n+1.$$ 
A: HINT
For the induction step assuming that $n^n>(n+1)!$ is true we have
$$(n+2)!=(n+2)(n+1)!\stackrel{Ind. Hyp.}<(n+2)n^n\stackrel{?}<(n+1)^{n+1}$$
thus we need to prove that 
$$(n+2)n^n<(n+1)^{n+1} \iff 1 + \frac1{n+1}<\left(1+\frac1n\right)^n$$
A: Hint: Note that$$\frac{(n+1)^{n+1}}{n^n}=(n+1)\left(1+\frac1n\right)^n$$and that$$\frac{(n+2)!}{(n+1)!}=n+2.$$It is not hard to prove that$$(\forall n\in\mathbb{N}):\left(1+\frac1n\right)^n>\frac{n+2}{n+1}.$$
A: With Bernoulli's inequality
$$\left(1+\dfrac{1}{n}\right)^n\geq1+n\dfrac{1}{n}=2>1+\dfrac{1}{n+1}$$
then
$$(n+1)^{n+1}>(n+2)n^n>(n+2)!$$
A: I'll start with the inductive hypothesis: 
$n^n>(n+1)!$
Then the inductive step: 
$(n+1)^{n+1}>(n+2)!$
I will prove this for 
$(n+1)^{n+1}>n^n(n+2)$ because $n^n(n+2) > (n+2)!$ given the inductive hypothesis. And therefore, if $(n+1)^{n+1}$ is  bigger than $n^n(n+2)$, it must also be bigger than $ (n+2)!$.
$(n+1)^{n+1}>n^n(n+2)$
$(n+1)^{n+1}>n^{n+1}+2n^n$
By binomial expansion, the first 2 terms of $(n+1)^{n+1}$ will be: 
$n^{n+1}$ and $n+1$ choose 1 multiplied to $n^n$
So we need to prove $n^{n+1}+(n+1)Choose1.......>n^{n+1}+2n^n$
We can simplify $(n+1)Choose1$ given the combinations formula: $$\frac{n!}{k!(n-k)!}$$ We know k=1 and n=n+1. This simplifies to n+1. And n+1 must be at least 4, since $n>=3$
Therefore even the first 2 terms of: $n^{n+1}+(n+1)n.....>n^{n+1}+2n^n$
This proves the statement. 
