Deriving Identity: A Proof for S^{p}_{n} = 1^p + ... + n^p

[quasar_4]

deriving identity - need help!

Homework Statement

Derive for S$$^{p}_{n}$$ = 1^p + ... + n^p the identity

(p+1)*S$$^{p}_{n}$$ + (p+1 choose 2)*S$$^{p-1}_{n}$$ + ...+S$$^{0}_{n}$$ = (n+1)^(p+1) - 1

Homework Equations

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

 

[quasar_4]

sorry, I don't know why this posted twice or how to delete the other one!

 

[Dick]

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?