prove that $\binom{k+1}{1} S_k(n)+\binom{k+1}{2} S_{k-1}(n)+.........+\binom{k+1}{k+1} S_0(n)=(n+1)^{k+1}-1$ For $n,k \in N$,we define
$S_k(n)=1^k +2^k+... +n^k$
if $(1+x)^p=1+\binom{p}{1}x+\binom{p}{2}x^2+....  +\binom{p}{p}x^p,p\in N$,then we have to prove that

$\binom{k+1}{1} S_k(n)+\binom{k+1}{2} S_{k-1}(n)+...+\binom{k+1}{k+1} S_0(n)=(n+1)^{k+1}-1$

this looks similar to $\binom{k+1}{1}n+\binom{k+1}{2}n^2+...+\binom{k+1}{k+1}n^{k+1}= (1+n)^{k+1}-\binom{k+1}{0}=(1+n)^{k+1}-1$
what next?
 A: By binomial theorem:
$$(1+x)^{m+1}-x^m=1+{n \choose 1}(x)+{n \choose 2} (x^2)+....+{m+1 \choose m} (x^m)$$
putting $x=1,2,3,,..n$ and adding them, we get
$$(n+1)^{n+1}-1=(1+1+1...+1)+{m+1 \choose 1}(1+2+3+...+n)+{m+1 \choose 2}(1^2+2^2+3^2++...+n^2)+{m+1 \choose 3} (1^3+2^3+3^3+4^3+...+n^3)+.....+{m+1 \choose m}(1^m+2^m+3^m+4^m+...+n^m)$$
$$\implies \sum_{k=0}^{m} S_k {m+1 \choose k}=(n+1)^{m+1}-1,~~ S_k=1^k+2^k+3^k+4^k+...n^k$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
\sum_{\ell = 1}^{k + 1}{k + 1 \choose \ell}\sum_{m = 1}^{n}m^{k -\ell + 1} & =
\sum_{m = 1}^{n}m^{k + 1}\sum_{\ell = 1}^{k + 1}{k + 1 \choose \ell}
\pars{1 \over m}^{\ell}
\\[5mm] & =
\sum_{m = 1}^{n}m^{k + 1}\bracks{\pars{1 + {1 \over m}}^{k + 1} - 1}
\\[5mm] & =
\sum_{m = 1}^{n}\pars{m + 1}^{k + 1} - \sum_{m = 1}^{n}m^{k + 1} =
\sum_{m = 2}^{n + 1}m^{k + 1} - \sum_{m = 1}^{n}m^{k + 1}
\\[5mm] & =
\bracks{-1 + \sum_{m = 1}^{n}m^{k + 1} + \pars{n + 1}^{k + 1}} -
\sum_{m = 1}^{n}m^{k + 1}
\\[5mm] & = \bbx{\pars{n + 1}^{k + 1} - 1}
\end{align}
