How to prove $ \sum_{k=1}^{p+1} \binom{p+1}{k}S_{n}^{p+1-k} = (n+1)^{p+1}-1$?

Question

This is the solution $ \sum_{k=1}^{p+1} \binom{p+1}{k}S_{n}^{p+1-k} = \sum_{k=1}^{p+1} \binom{p+1}{k} \sum_{l=1}^{n}l^{p+1-k} = \sum_{l=1}^{n}(l+1)^{p+1}-l^{p+1} = (n+1)^{p+1}-1$ ?

With $S_{n}^{p} = \sum_{k=1}^{n}k^p$.

Answer 1

Reverse the order of summation and apply the binomial theorem:

$$\begin{align*} \sum_{k=1}^{p+1}\binom{p+1}k\sum_{\ell=1}^n\ell^{p+1-k}&=\sum_{\ell=1}^n\sum_{k=1}^{p+1}\binom{p+1}k\ell^{p+1-k}\\\\ &=\sum_{\ell=1}^n\sum_{k=1}^{p+1}\binom{p+1}k1^k\ell^{p+1-k}\\\\ &=\sum_{\ell=1}^n\left(\sum_{k=0}^{p+1}\binom{p+1}k1^k\ell^{p+1-k}-\binom{p+1}01^0\ell^{p+1}\right)\\\\ &=\sum_{\ell=1}^n\left((\ell+1)^{p+1}-\ell^{p+1}\right)\\\\ &=(n+1)^{p+1}-1\;. \end{align*}$$