Let $n \in \mathbb{Z}^+$, prove the identity $ \sum_{k=1}^{n-1} \binom {n} {k} \frac{kn^{n-k}}{k+1}=\frac{n(n^{n}-1)}{n+1}$ Let $n \in \mathbb{Z}^+$, prove the identity $$ \sum_{k=1}^{n-1} \binom {n} {k} \frac{kn^{n-k}}{k+1}=\frac{n(n^{n}-1)}{n+1}$$
First of all  $$ \sum_{k=1}^{n-1} \binom {n} {k} \frac{kn^{n-k}}{k+1}=n^{n}\Bigg(\sum_{k=1}^{n-1} \binom {n} {k} \frac{kn^{-k}}{k+1} \Bigg)$$
$$=n^n\Bigg(\sum_{k=1}^{n-1} \binom {n}{k} \bigg(1-\frac{1}{k+1}\bigg)\bigg(\frac{1}{n^k}\bigg)\Bigg)$$
$$=n^n \sum_{k=1}^{n-1} \binom {n} {k} \frac{1}{n^k}-n^n \sum_{k=1}^{n-1} \binom {n}{k}  \bigg( \frac{1}{k+1}\bigg)\bigg(\frac{1}{n^k} \bigg)$$
We have for the first sum $$(1+\frac{1}{x})^n = \sum_{k=0}^n \binom{n}{k}\frac{1}{x^k}.$$
For the second sum 
$$(1+x)^n = \sum_{k=0}^n \binom{n}{k}x^k.$$
Integrating both sides from $0$ to $x$, we see that 
$$\frac{(1+x)^{n+1}-1}{n+1} = \sum_{k=0}^n \binom{n}{k}\frac{x^{k+1}}{k+1}.$$
Putting $x=1$ yields $$\frac{2^{n+1}-1}{n+1}=\sum_{k=0}^n \binom{n}{k}\frac{1}{k+1}$$
Here where I have stopped. I could not get them similar for what I have. Would someone help me out !

$$
\begin{align}
\sum_{k=1}^n\binom{n}{k}\frac{kn^{n-k}}{k+1}
&=n^n\sum_{k=1}^n\binom{n-1}{k-1}\frac1{k+1}\frac1{n^{k-1}}\\
&=n^n\int_0^1\sum_{k=1}^n\binom{n-1}{k-1}\frac{t^k}{n^{k-1}}\,\mathrm{d}t\\
&=n^n\int_0^1t\sum_{k=0}^{n-1}\binom{n-1}{k}\frac{t^k}{n^k}\,\mathrm{d}t\\
&=n^n\int_0^1t\,\left(1+\frac tn\right)^{n-1}\,\mathrm{d}t\\
&=n\int_0^1t\,(n+t)^{n-1}\,\mathrm{d}t\\
&=n\int_n^{n+1}(\color{#C00000}{t}-\color{#00A000}{n})\,t^{n-1}\,\mathrm{d}t\\
&=\color{#C00000}{\frac{n(n+1)^{n+1}-n^{n+2}}{n+1}}-\color{#00A000}{\left(n(n+1)^n-n^{n+1}\right)}\\
&=\frac{n^{n+1}}{n+1}\tag{1}
\end{align}
$$
Subtract the $k=n$ term from $(1)$ and we get
$$
\sum_{k=1}^{n-1}\binom{n}{k}\frac{kn^{n-k}}{k+1}=\frac{n^{n+1}-n}{n+1}\tag{2}
$$
