Deriving sum of powers formula using generating functions 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 $H(x,p)$?
 A: It's convenient to use the  differential    operator $D_x:=\frac{d}{dx}$    and   the coefficient of operator $[x^n]$     to     denote     the coefficient  of  $x^n$  of a  series $A(x)=\sum_{k=0}^\infty a_kx^k$.

Hint: Study the operator
  \begin{align*}
[x^n]\frac{1}{1-x}(xD_x)^p
\end{align*}
  applied on $\frac{1}{1-x}$.

A detailed answer can be found here.
A: This answer may not follow the same line of reasoning but you might want to try out using derivatives and integrals of the generating function of $x^n$. the derivative of $1+x+x^2...+x^s$ is $1+2x+3x^2...sx^{s-1}$. On the rhs, we can differentiate the generating functions of the corresponding series. The derivative of $\frac{1-x^{s+1}}{1-x} = \frac{1-(1+s)x^s + s x^{1+s}}{(1-x)^2}$. Then apply l'Hopital's rule as $lim_{x \to\ 1}$. That will give you a function for the sum up to $s$. It is easy to see that in order to find the sum of $p$ powers, one must multiply the result by $x$ and differentiate the rhs, $p$ times. 
Further, if we want to find the sum of inverse powers using generating functions, we must integrate the geometric series generating function and divide by $x$, $p$ times. This will not however give us a 'closed form' solution of formula in the case of inverse powers unless you consider the formula for the infinite sum of harmonic numbers, which is $-ln(x-1)$, this is technically true since the harmonic series ends up growing like the natural logarithm function. Also, this formula is useful because if you work it out for $x=-1$ you get the infinite sum of the alternating harmonic numbers - ln(2) - which is correct although the series is only conditionally convergent.
The method is pretty cumbersome since differentiating the fractions becomes increasingly complicated and you will have to apply l'hopital's rule $p+1$ times on the resulting fraction. So generating functions, (at least the method I used) would not really be practical to carry out unless you use software. A better method is described here...
https://en.wikipedia.org/wiki/Faulhaber%27s_formula
A: If I may, I will use the derivative but only very briefly.
Consider:
$$\frac {d^p}{dx^p}e^{kx}=k^pe^{kx}$$
Now sum over $k$:
$$\sum_{k=0}^n \frac {d^p}{dx^p}e^{kx} =\frac {d^p}{dx^p}\sum_{k=0}^n e^{kx}$$
You will recognize the geometric series on the right. This yields:
$$G(x)=\frac {e^{(n+1)x}-1}{e^x-1}$$
