<s>[INST] 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?
 [/INST] 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
$$(n+1)^{n+1}-1=(1+1+1...+1)+{m+1 \choose 1}(1+2+3+...+n)+{m+1 \choose 2}(1^2+2^2+3^2++...+n^2)+{m+1 \choose 3} (1^3+2^3+3^3+4^3+...+n^3)+.....+{m+1 \choose m}(1^m+2^m+3^m+4^m+...+n^m)$$
$$\implies \sum_{k=0}^{m} S_k {m+1 \choose k}=(n+1)^{m+1}-1,~~ S_k=1^k+2^k+3^k+4^k+...n^k$$
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