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. 

Try taking log of both sides and prove $\frac{\log x}x$ is decreasing.
Or by induction try to show $(\frac{k+1}k)^k\leq k$:
$$(1+1/k)^k\leq \sum_{i=0}^k \binom ki k^{-i}<\sum_{i=0}^k 1=k+1$$
