$$ \text{For the general power sum: } S_n^p:=1^p+2^p+3^p...n^p $$ $$ \text{Show that: }(p+1)S_n^p + {p+1\choose 2}S_n^{p-1} + {p+1\choose 3}S_n^{p-2}+...+S_n^0=(n+1)^{p+1}-1 $$
A part of the answer is:
$$ (p+1)S_n^p + {p+1\choose 2}S_n^{p-1} + {p+1\choose 3}S_n^{p-2}+...+S_n^0=(p+1)[1^p+2^p+...+n^p]+{p+1\choose2}[1^{p-1}+2^{p-1}+...+n^{p-1}]+...+1*[1^0+2^0+...+n^0]$$$$=\sum_{k=0}^{p}1^k{p+1 \choose k}+\sum_{k=0}^{p}2^k{p+1 \choose k}+...+\sum_{k=0}^{p}n^k{p+1 \choose k} $$ I cant understand how come that when the first two elements of the sum are: $$ \text{I.} (p+1)S_n^p + {p+1\choose 2}S_n^{p-1}... $$ It can be rewritten to: $$ \text{II.}\sum_{k=0}^{p}1^k{p+1 \choose k}+\sum_{k=0}^{p}2^k{p+1 \choose k}{...} $$ and so on.. How come that in II we can have k=1 but in I k starts as 0 and afterwards it is 2 and so on?
