What is this polynomial? I came across the "summation by parts" formula, and wondered how it could be useful in any way:
$$
\sum_{k=n}^m\ a_k \Delta b_k=\left[a_k b_k\right]_{n}^{m+1}-\sum_{k=n}^{m}\ b_{k+1} \Delta a_k
$$
Where $\Delta a_k$ is the Forward difference operator. I tried to do the following: with $|\alpha|<1$:
$$
\sum_{k=0}^m \ k \alpha^k=\frac{1}{\alpha-1}\sum_{k=0}^m \ k \alpha^k(\alpha-1)=\frac{1}{\alpha-1}\left(\left[ k\alpha^k\right]_0^{m+1}-\sum_{k=0}^m\ \alpha^{k+1}\right)
$$
And then I took the limit as  $m\to \infty$, getting:
$$
\sum_{k=0}^\infty\ k\alpha^k=\frac{\alpha}{(1-\alpha)^2}
$$
Which is a result I'm sure many of you know, but then I realized that, with this formula, i could do the same for $\sum_{k=0}^\infty\ k^2 \alpha^k$, being $\Delta k^2=2k+1$, arriving at:
$$
\sum_{k=0}^\infty\ k^2 \alpha^k=\frac{\alpha}{(1-\alpha)^2}\left(2\cdot\frac{\alpha}{1-\alpha}+1\right)
$$
But I could do the same for $\sum_{k=0}^{\infty}\ k^3\alpha^k$, and so on, obtaining this formula:
$$
\sum_{k=0}^\infty\ k^3\alpha^k=\frac{\alpha}{(1-\alpha)^2}\left(6\cdot \left(\frac{\alpha}{1-\alpha}\right)^2+6\cdot \left(\frac{\alpha}{1-\alpha}\right)+1\right)
$$
And so on with the fourth degree and the fifth:
$$
\sum_{k=0}^{\infty}\ k^4\alpha^k=\frac{\alpha}{(1-\alpha)^2}\left(24\cdot \left(\frac{\alpha}{1-\alpha}\right)^3+36\cdot\left(\frac{\alpha}{1-\alpha}\right)^2+14\cdot\left(\frac{\alpha}{1-\alpha}\right)+1\right)
$$
$$
\sum_{k=0}^{\infty}\ k^5\alpha^k=\frac{\alpha}{(1-\alpha)^2}\left(120\cdot\left(\frac{\alpha}{1-\alpha}\right)^4+240\cdot \left(\frac{\alpha}{1-\alpha}\right)^3+150\cdot\left(\frac{\alpha}{1-\alpha}\right)^2+30\cdot \left(\frac{\alpha}{1-\alpha}\right)+1\right)
$$
At this point I got bored and wanted to find a general formula, seeing a polynomial pattern, but all i could figure out was this: (I didn't prove any of the following, I just tried to figure out the patterns, which could be misleading)
Given $|\alpha|<1$, and $n\in \mathbb{N}$
$$
\sum_{k=0}^\infty k^n \alpha^k=\frac{\alpha}{(1-\alpha)^2}\left(\sum_{k=0}^{n-1} P_{n,k} \left(\frac{\alpha}{1-\alpha}\right)^k\right)
$$
Where $P_{n,k}$ is a coefficient defined for $n\in \mathbb{N},\ k\in \{0,1,2,...,n-1\}$, with these defining properties:
$$
\forall n \in \mathbb{N}, P_{n,0}=1\\
P_{n,k}=\sum_{h=1}^{n-k}\binom{n}{h}P_{n-h,k-1}
$$
I'm having some troubles on how to get to a closed form of theese coefficients, all i know is that the coefficient of the highest term is $n!$, the known term is $1$, and that the linear term is $2^n-2$, the formulas for the other coefficients are a mess beyond my comprehension, with nested sums of binomial coefficients. If you recognize these polynomials by any chance, or if you can find a closed form for them, or any kind of insight on this, please let me know.
Post Scriptum: I didn't prove any of the results i showed about $P_{n,k}$, so if you got a correction to make, please point that out.
 A: The results you obtained, say for $n=3$, are based on the fact that
$$k^3=k(k-1)(k-2)+3k(k-1)+k$$
So
$$\sum_{k=0}^\infty k^3 a^k=\sum_{k=0}^\infty k(k-1)(k-2) a^k+3\sum_{k=0}^\infty k(k-1)a^k+\sum_{k=0}^\infty ka^k$$ that is to say
$$a^3\sum_{k=0}^\infty k(k-1)(k-2) a^{k-3}+3a^2\sum_{k=0}^\infty k(k-1)a^{k-2}+k\sum_{k=0}^\infty ka^{k-1}$$ that is to say
$$\sum_{k=0}^\infty k^3 a^k=a^3\left(\sum_{k=0}^\infty a^{k} \right)'''+3a^2\left(\sum_{k=0}^\infty a^{k} \right)''+a\left(\sum_{k=0}^\infty a^{k} \right)'$$
Now, for the general case of
$$S_n=\sum_{k=0}^\infty k^n a^k=\Phi (a,-n,0)$$ where appears the Hurwitz-Lerch transcendent function.
This can write
$$S_n=\frac a {(1-a)^{n+1}}P_{n-2}(a)$$
The first polynomials are
$$\left(
\begin{array}{cc}
 n & P_n(a) \\ 
 1 & 1 \\
 2 & a+1 \\
 3 & a^2+4 a+1 \\
 4 & a^3+11 a^2+11 a+1 \\
 5 & a^4+26 a^3+66 a^2+26 a+1 \\
 6 & a^5+57 a^4+302 a^3+302 a^2+57 a+1 \\
 7 & a^6+120 a^5+1191 a^4+2416 a^3+1191 a^2+120 a+1 \\
 8 & a^7+247 a^6+4293 a^5+15619 a^4+15619 a^3+4293 a^2+247 a+1
\end{array}
\right)$$ where you can first notice that they are palindromic.
The  coefficientcorresponds to  Eulerian numbers (just seach for them in $OEIS$)
