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.

We are required to prove that: $(k+1)^{k+2}>(k+1)(k+2)^k$.
The assumption is that: $k^{k+1} > k(k+1)^{k-1}$. Dividng both sides by $k^k$ gives: $k> \left(\frac{k+1}{k}\right)^{k-1}$, thus $k>\left(1+\frac{1}{k}\right)^{k-1}$
Starting from this, 
\begin{eqnarray}
k&>&\left(1+\frac{1}{k}\right)^{k-1}\\
\frac{1}{\left(1+\frac{1}{k}\right)^{k-1}}&>&\frac{1}{k}\\
1+\frac{1}{\left(1+\frac{1}{k}\right)^{k-1}}&>&1+\frac{1}{k}\\
\left(1+\frac{1}{k}\right)^{k-1}+1&>&\left(1+\frac{1}{k}\right)^k\\
\end{eqnarray}
Now, since $k>\left(1+\frac{1}{k}\right)^{k-1}$, then: 
\begin{eqnarray}
k+1&>&\left(1+\frac{1}{k}\right)^k\\
(k+1)^{k+2}&>&(k+1)^{k+1}\left(1+\frac{1}{k}\right)^k
\end{eqnarray}
Also, $\frac{1}{k}>\frac{1}{k+1}$, therefore:
\begin{eqnarray} 
(k+1)^{k+2}&>&(k+1)^{k+1}\left(1+\frac{1}{k+1}\right)^k\\
&>&(k+1)(k+2)^k
\end{eqnarray}
Therefore, $(k+1)^{k+2}>(k+1)(k+2)^k$. $\Box$
