Summation of nth-degree nos.

Denote Sk (n)=1^k + 2^k + ... + n^k. There exist a recursive formula such that you can use the previous formula of Sk-1 (n), Sk-2 (n),... S1 (n) to find out Sk (n)
k+1C1Sk (n)+ k+1C2Sk-1 (n)+...+k+1Ck+1S0 (n)=(n+1)k+1-1
where S0 (n)=n.
For example, 2C1S1 (n)+2C2S0 (n)=(n+1)2-1
S1 (n)=(n2+2n-n)/2=n(n+1)/2
There is a similar formula which is related to Bernoulli numbers. However, since to find out those Bernoulli numbers also requires using other recursive formulae. So, I omit them here.