Deriving identity for power sum S^p_n using binomial coefficients

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deriving identity - need help!

Homework Statement



Derive for S[tex]^{p}_{n}[/tex] = 1^p + ... + n^p the identity

(p+1)*S[tex]^{p}_{n}[/tex] + (p+1 choose 2)*S[tex]^{p-1}_{n}[/tex] + ...+S[tex]^{0}_{n}[/tex] = (n+1)^(p+1) - 1

Homework Equations



Um, I know that the S[tex]^{1}_{n}[/tex] = n(n+1)/2
S[tex]^{2}_{n}[/tex] = n(n+1)(2n+1)/6
S[tex]^{3}_{n}[/tex] = [1+2+...+n]^2

The Attempt at a Solution



I have NO idea how to show this. I tried writing out some of the terms, but I didn't really get anywhere. I am completely lost as to how my lhs is supposed to become (n+1)^ anything... yeah... all I know is that I can write out the p choose n kind of terms, but so far that hasn't really yielded anything useful. Please help! I am so confused! :cry:
 
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sorry, I don't know why this posted twice or how to delete the other one!
 


I'll give you a hint. Write C(n,m) for "n choose m". Now the sum for k=0 to n C(n,i)*k^i is (1+k)^n, right? So the sum for k=0 to n-1 of C(n,i)*k^i is (1+k)^n-k^n. If you sum over k, do you see a telescoping series?