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.

From the definition of $P_k(n)$ we get
$$
P_k(n) = \sum_{i=1}^n i^k \Rightarrow 
P_0(n) = \sum_{i=1}^n 1 = n.
$$
We now use complete induction over $k$ to proof the statement $S(k)$
$$
P_k(x) = k \int\limits_0^x P_{k-1}(t) \, dt + C_k \, x \quad (*)
$$
Base case
For $k=1$ we have $S(1)$:
$$
1 \int\limits_0^x P_0(t) \, dt + C_1 \, x 
=
\int\limits_0^x t \, dt + C_1 \, x
=
\frac{1}{2} x^2 + C_1 \, x 
$$
which corresponds to the well known Gauss summation formula $P_1(n)$, if $C_1 = 1/2$.
Inductive step
Assuming equation $(*)$ is true for $\{ 1, \ldots, k \}$ we perform the recursion by starting with $S(k)$ and then applying $S(k-1), S(k-2), \ldots$ until we hit the bottom with $S(1)$ and get $P_0 = \mbox{id}$. That integrand is then simple enough to perform the the $k$ iterated integrations:
\begin{align}
P_k(x) 
&= 
k \int\limits_0^x P_{k-1}(t_k) \, dt_k + C_k \, x \\
&=
k \int\limits_0^x \left( (k-1) \int\limits_0^{t_k} 
P_{k-2}(t_{k-1}) \, dt_{k-1} + C_{k-1} \, t_k \right) \, dt_k + C_k \, x  \\
&=
k (k-1) \int\limits_0^x \int\limits_0^{t_k} 
P_{k-2}(t_{k-1}) \, dt_{k-1} \, dt_k + \frac{k}{2} C_{k-1} \, x^2 + C_k \, x  \\
&=
k! \int\limits_0^x \cdots \int\limits_0^{t_2} 
P_0(t_1) \, dt_1 \cdots \,dt_k
+ \sum_{j=1}^k C_{k-j+1} \frac{k!}{(k-j+1)!j!} x^j \\
&=
k! \int\limits_0^x \cdots \int\limits_0^{t_2} 
t_1 \, dt_1 \cdots \,dt_k
+ \sum_{j=1}^k \binom{k}{j} \frac{C_{k-j+1}}{k-j+1} x^j \\
&=
k! \int\limits_0^x \cdots \int\limits_0^{t_3}
\frac{1}{2}t_2^2 \, dt_2 \cdots \,dt_k
+ \sum_{j=1}^k \binom{k}{j} \frac{C_{k-j+1}}{k-j+1} x^j \\
&=
k! \int\limits_0^x
\frac{1}{k!}t_k^k \,dt_k
+ \sum_{j=1}^k \binom{k}{j} \frac{C_{k-j+1}}{k-j+1} x^j \\
&=
\frac{1}{k+1}x^{k+1} 
+ \sum_{j=1}^k \binom{k}{j} \frac{C_{k-j+1}}{k-j+1} x^j \\
\end{align}
Then we try to arrive at $S(k+1)$:
\begin{align}
(k+1) \int\limits_0^x P_k(t) \, dt + C_{k+1} \, x
&=
(k+1) \int\limits_0^x \left(
\frac{1}{k+1}t^{k+1} + 
\sum_{j=1}^k \binom{k}{j} \frac{C_{k-j+1}}{k-j+1} t^j \right)
\, dt \\
& + C_{k+1} \, x \\
&=
(k+1) \left(
\frac{1}{(k+1)(k+2)}x^{k+2} 
+ \right. \\
& \left. \sum_{j=1}^k \binom{k}{j} \frac{C_{k-j+1}}{(k-j+1)(j+1)} x^{j+1}
\right) + C_{k+1} \, x \\
&=
\frac{1}{k+2}x^{k+2} 
+ \sum_{j=1}^k \binom{k}{j} \frac{(k+1)C_{k-j+1}}{(k-j+1)(j+1)} x^{j+1}
+ C_{k+1} \, x \\
&=
\frac{1}{k+2}x^{k+2} 
+ \sum_{j=1}^k \binom{k+1}{j+1} \frac{C_{k-j+1}}{k-j+1} x^{j+1}
+ C_{k+1} \, x \\
&=
\frac{1}{k+2}x^{k+2} 
+ \sum_{j=2}^{k+1} \binom{k+1}{j} \frac{C_{k+1-j+1}}{k+1-j+1} x^j
+ C_{k+1} \, x \\
&=
\frac{1}{k+2}x^{k+2} 
+ \sum_{j=1}^{k+1} \binom{k+1}{j} \frac{C_{k+1-j+1}}{k+1-j+1} x^j \\
&=
P_{k+1}(x)
\end{align}
Thus $S(k+1)$ follows.
By the principle of induction $(*)$ holds for all $k \in \mathbb{N} \setminus \{ 0\}$.
