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.
 A: 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, $(k+1)^{k+2}>(k+1)(k+2)^k$. $\Box$
A: At step 3: when n = k + 1 the inequality becomes: (k+1)^(k+2) > (k+1)*(k+2)^k <===>
(k+1)^(k+1) > (k+2)^k <===> (k+1)^(k+1) > ((k+1) + 1)^k <===> k+1 > (1 + 1/(k+1))^k ) (*).
(*) is true because:


*

*k+1 > 3

*3 > (1 + 1/(k+1))^(k+1) ( reminder: the limit of the right side is e = 2.71728 < 3 )

*(1 + 1/(k+1))^(k+1) > (1 + 1/(k+1))^k
A: Define: $\rm f(n) = \dfrac{n^{n+1}}{n(n+1)^{n-1}} \implies \dfrac{f(n+1)}{f(n)} = \dfrac{(n+1)^{2n}}{n^n(n+2)^n} = \left(\dfrac{n^2 + 2n + 1}{n^2 + 2n}\right)^n > 1$
Base Case: $\rm f(2) = \dfrac{2^3}{2 \cdot 3^1} = \dfrac{8}{6} > 1$
Inductive Step: $\rm f(n) > 1 \implies f(n+1) = \underbrace{\left(\dfrac{f(n+1)}{f(n)}\right)}{} \cdot \underbrace{f(n)}{} > 1$ since  both terms in the product are greater than $1$ individually.
See here and here for other examples of induction using multiplicative telescopy, a very powerful technique illuminated by Bill Dubuque.
A: 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
$\large (k+1)^{k+2}>(k+1)(k+2)^k$
