Sum of $k$-th powers Given:
$$
P_k(n)=\sum_{i=1}^n i^k
$$
and $P_k(0)=0$, $P_k(x)-P_k(x-1) = x^k$ show that:
$$
P_{k+1}(x)=(k+1) \int^x_0P_k(t) \, dt + C_{k+1} \cdot x
$$
For $C_{k+1}$ constant.
I believe a proof by induction is the way to go here, and have shown the case for $k=0$. This is where I'm stuck. I have looked at the right hand side for the k+1 case:
$$
(k+2)\int^x_0P_{k+1}(t) \, dt + C_{k+2} \cdot x
$$
and I don't see how this reduces to $P_{k+2}(x)$. Even if we are assuming the kth case, replacing $P_{k+1}$ in the integrand of the $(k+1)$-st case just makes it more messy. I am not looking for the answer just a push in the right direction. I can see that each sum ends up as a polynomial since expressions like $P_1(x) = 1+2+\cdots+x=\frac{x(x+1)}{2}$, but I don't know how to do that for arbitrary powers, and I believe I don't need to in order to solve this problem.

Hm. This problem is weird, in that as defined $P_k(n)$ is only defined on the naturals. Although as you noted, you can find a closed form and evaluate it at an arbitrary point.
Have you tried taking the derivative of both sides and using Fundamental Thm of Calculus? That would get rid of the integral, and turns the $C_{k+1}\cdot x$ term to just a constant, which looks more appealing.
