Show that $(n+1)^{n+1}>(n+2)^n$ for all positive integers

Question

Show that:

$(n+1)^{n+1}>(n+2)^n$ holds for all positive integers

I tried using induction:

for $n=1$ we have 4>3 then for $n+1$ we have to show that $(n+2)^{n+2}>(n+3)^{n+1}$ and here I stuck

Answer 1

Writing it as $m^m>(m+1)^{m-1}$ for $m>1$ might make it a little easier.

First, show $\binom{m-1}{i}\leq m^{m-1-i}$, with strict inequality when $0\leq i<m-1$.

Then, for any positive $m$, $$(m+1)^{m-1}=\sum_{i=0}^{m-1}\binom{m-1}{i} m^i\leq \sum_{i=0}^{m-1} m^{m-1} = m^m$$ with strict inequality when $m>1$.

Answer 2

Since $(1+x)^n \geq 1 + nx, \forall x\geq -1$

$(\dfrac{n+1}{n+2})^n=(1-\dfrac{1}{n+2})^n \geq 1 -\dfrac{n}{n+2} = \dfrac{2}{n+2}$

$(n+1)\dfrac{2}{n+2} >1$