<s>[INST] Induction proof of $n^{(n+1) }> n(n+1)^{(n-1)}$ The question statement from my homework booklet goes:

Prove by mathematical induction that $n^{n+1} > n(n+1)^{n-1}$ 
  is true for all integers $n \geq 2$.

I've managed to come up with this for the induction step (the base case is trivial), but I am not sure what to do from here:
Assume true for n=k. For n=k+1,
\begin{align*}
&k^{k+1} > k(k+1)^{k+1} \\
&(k+1)k^{k+1} > k(k+1)^k \\
&(k+1)^{k+2} < (k+1)^3 k^k < (k+1)^3 (k+2)^k \\
&(k+1)^{k-1} < (k+2)^k
\end{align*}
I would greatly appreciate any help with how to solve this. Thanks in advance.
 [/INST] For $k$ we have
$\large k^{k+1}>k(k+1)^{k-1}$
which we hold to be true.
For $k+1$ we have
$\large(k+1)^{k+2}=(k+1)^{k+1}(k+1)=k^{k+1}(1+\frac{1}{k})^{k+1}(k+1)
\\\large>k(1+k)^k(1+\frac{1}{k})^{k+1}$
as $k^{k+1}>k(k+1)^{k-1}$
Now 
$\large k(k+1)^k(1+\frac{1}{k})^{k+1}=k(k+1)^k(1+\frac{1}{k})^{k+1}(\frac{k+2}{k+2})^k
\\\large =k(k+1)^k(\frac{k+1}{k})^{k+1}(\frac{k+2}{k+2})^k=(\frac{(k+1)^2}{k(k+2)})^k(k+1)(k+2)^k\\\large>(k+1)(k+2)^k$
where we have used the fact that
$(k+1)^2=k(k+2)+1>k(k+2)$
so that
$\large (\frac{(k+1)^2}{k(k+2)})^k>\frac{(k+1)^2}{k(k+2)}>1$
Thus we have
$\large (k+1)^{k+2}>k(1+k)^k(1+\frac{1}{k})^{k+1}>(k+1)(k+2)^k$
Culminating in
$\large (k+1)^{k+2}>(k+1)(k+2)^k$
</s>