Just for fun I wanted to try to derive a formula for the sum of $p$-powers using generating functions, but without using any literature or websites for help. However I do need a small push or hint.
Let $p$ be some positive integer constant.
Define $f(n) = \sum_{k=0}^{n} k^p = 0^p + 1^p + 2^p + 3^p + \cdots + n^p$
I can also state it as a recurrence: $f(n) = f(n-1) + n^p$ where $f(0) = 0$.
Define $G(x) = \sum_{n=0}^{\infty} f(n) x^n$
Then
$$ \begin{aligned} G(x) &= f(0)x^0 + \sum_{n=1}^{\infty} f(n) x^n\\ G(x) &= 0 + \sum_{n=1}^{\infty} f(n-1) x^n + \sum_{n=1}^{\infty} n^p x^n\\ G\left( x \right) &=x\sum_{n=1}^{\infty}{f}\left( n-1 \right) x^{n-1}+\left( -0^px^0+\sum_{n=0}^{\infty}{n}^px^n \right) \\ G(x) &= x\sum_{n=0}^{\infty} f(n) x^{n} + \sum_{n=0}^{\infty} n^p x^n\\ G(x) &= xG(x) + \sum_{n=0}^{\infty} n^p x^n\\ G(x) - xG(x) &= \sum_{n=0}^{\infty} n^p x^n\\ G(x) &= \frac{\sum_{n=0}^{\infty} n^p x^n}{1-x} \end{aligned}$$
So now it is all about finding the generating function for $H(x,p) = \sum_{n=0}^{\infty} n^p x^n$
I need some way to get from $H(x,p-1) = \sum_{n=0}^{\infty} n^{p-1} x^n$ to $H(x,p) = \sum_{n=0}^{\infty} n^p x^n$ because the base case is $H(x,0) = \sum_{n=0}^{\infty} n^0 x^n = \sum_{n=0}^{\infty} x^n = \frac{1}{1-x}$
At this point I feel a little stuck and could use a push in the right direction. Am I onto a solution here or am I just spinning my wheels? Where can I go from here? I know one usual approach is to keep taking the derivative of both sides but I'd prefer to avoid that method (no real reason, just want to see if it can be done without noticing that trick).
How can I relate $H(x,p-1)$ to $H(x,p)$?
