<s>[INST] 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 !
 [/INST] There is a straightforward direct computational argument:
$$\begin{align*}
\sum_{k=1}^{n-1}\binom{n}k\frac{kn^{n-k}}{k+1}&=\sum_{k=1}^{n-1}\binom{n}k\left(1-\frac{1}{k+1}\right)n^{n-k}\\
&=\sum_{k=1}^{n-1}\binom{n}kn^{n-k}-\sum_{k=1}^{n-1}\binom{n}k\frac{n^{n-k}}{k+1}\\
&=(n+1)^n-n^n-1-\frac{1}{n+1}\sum_{k=1}^{n-1}\binom{n+1}{k+1}n^{n-k}\\
&=(n+1)^n-n^n-1-\frac{1}{n+1}\sum_{k=2}^n\binom{n+1}kn^{n+1-k}\\
&=(n+1)^n-n^n-1-\frac{1}{n+1}\left((n+1)^{n+1}-n^{n+1}-(n+1)n^n-1\right)\\
&=-1+\frac{n^{n+1}+1}{n+1}\\
&=\frac{n(n^n-1)}{n+1}
\end{align*}$$
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