I have to prove this identity. I tried with induction and Newton's binomial, but I get stuck at some point. Could you tell me please what to do? S(k, n) = 1^k + 2^k +...+ n^k (you can see the identity by clicking on the sum, I'm sorry, I'm new to this site!) $$S_k(n)=\sum _{m=1}^n m^k$$ $$1+\sum _{k=0}^{r-1} S_k(n) \binom{r}{k}=(n+1)^r$$
And after that I have to find the formula for S(1, n), S(2, n), S(3, n) and S(4, n). Thank you in advance!
Hint:
Write successively \begin{align} (n+1)^k&= n^k+\binom k1 n^{k-1}+\binom k2 n^{k-2}+\dots+\binom k{k-1}n+1 \\ n^k=(n-1+1)^k&= (n-1)^k+\binom k1 (n-1)^{k-1}+\binom k2(n-1)^{k-2}+\dots+\binom k{k-1}(n-1)+1 \\ \vdots \\ 2^k=(1+1)^k&= 1^k+\binom k1 1^{k-1}+\binom k2 1^{k-2}+\dots+\binom k{k-1}1+1 , \end{align} add all these equations and simplify.