Prove by induction that $(k + 2)^{k + 1} \leq (k+1)^{k +2}$ 
Prove by induction that $$ (k + 2)^{k + 1} \leq (k+1)^{k +2}$$ for $ k > 3 .$

I have been trying to solve this, but I am not getting the sufficient insight. 
For example,  $(k + 2)^{k + 1} = (k +2)^k (k +2) , (k+1)^{k +2}= (k+1)^k(k +1)^2.$
$(k +2) < (k +1)^2 $ but  $(k+1)^k < (k +2)^k$ so what I want would clearly not be immediate from using something like If  $ 0 < a < b, 0<c<d $ then $0 < ac < bd $. THe formula is valid for n = 4, So if it is valid for $n = k$ I would have to use
$ (k + 2)^{k + 1} \leq (k+1)^{k +2} $ somewhere in order to get that $ (k + 3)^{k + 2} \leq (k+2)^{k +3} $ is also valid. This seems tricky.
I also tried expanding  $(k +2)^k $ using the binomial formula and multiplying this by $(k + 2)$, and I expanded $(k+1)^k$ and multiplied it by $(k + 1)^2 $ term by term. I tried to compare these sums, but it also gets tricky. I would appreaciate a hint for this problem, thanks. 

A simpler (and a little stronger) statement is:
$$\forall n\ge3\quad(n+1)^n<n^{n+1}.$$
We first check that $(3+1)^3=64<81=3^{3+1}.$
Then, for the induction step, it is sufficient to prove (for all $n\ge3$) that
$$\frac{(n+2)^{n+1}}{(n+1)^{n+2}}\le\frac{(n+1)^n}{n^{n+1}},$$
i.e.
$$(n(n+2))^{n+1}\le(n+1)^{2n+2},$$
i.e.
$$n(n+2)\le(n+1)^2,$$
which is obvious.
