Show that $n^n>(n+1)!$ for all $n\ge3$

Question

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.

Answer 1

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$$(\forall n\in\mathbb{N}):\left(1+\frac1n\right)^n>\frac{n+2}{n+1}.$$

Answer 2

With Bernoulli's inequality $$\left(1+\dfrac{1}{n}\right)^n\geq1+n\dfrac{1}{n}=2>1+\dfrac{1}{n+1}$$ then $$(n+1)^{n+1}>(n+2)n^n>(n+2)!$$

Answer 3

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

$$(n+2)n^n<(n+1)^{n+1} \iff 1 + \frac1{n+1}<\left(1+\frac1n\right)^n$$

Answer 4

$$\frac{n^n}{n!}=\frac n1\cdot\frac n2\cdot\frac n3\cdots\frac nn>\frac n1\cdot\frac n2>n+1.$$

Answer 5

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.