Answer
See proof (use induction)
Work Step by Step
Consider the equality:
$A(n)$: $P(n)=1^k+2^k+... +n^k$
We will use induction to prove that $A(n)$ is true.
Step 1: $n=1$
$P(1)=P(1+0)=P(0+1)-P(0)$ (because $P(0)=0$
$=\delta P(0)$
$=(0+1)^k$
$=1^k$
Therefore $A(1)$ is true.
Step 2: Assume that $A(r)$ is true:
$A(r)$: $P(r)=1^k+2^k+....+r^k$
$\delta P(r)=(r+1)^k$
$P(r+1)-P(r)=(r+1)^k$
$P(r+1)=P(r)+(r+1)^k$
$=1^k+2^k+....+r^k+(r+1)^k$
Therefore $A(r+1)$ is true.
So $A(n)$ is true for all $n$.
