To make manipulations more tractable, I'll consider $S_N(x)=\sum_{i=0}^\infty i^N x^i$ for $N=0,1,2$ and leave the substitution of $x=1/4$ to the reader. For $N=0$ we have the usual geometric sum $S_0(x)=\dfrac{1}{1-x}$. The familiar proof of this can be boiled down to observing that
\begin{align}
(1-x)S_0(x)=S_0(x)-x S_0(x)
=1&+x+x^2+\cdots\\
&-x-x^2-\cdots=1.
\end{align}
This contains an insight which generalizes to any given power series $A(x)=\sum_{i=0}^\infty a_i x^i$:
\begin{align}
(1-x)A(x)=A(x)-x A(x)
=a_0&+a_1 x + a_2 x^2+\cdots\\
&-a_0x-a_1x^2-\cdots=a_0+\sum_{i=1}^\infty (a_{i+1}-a_i)x^i
\end{align}
In other words, multiplying by $(1-x)$ replaces the coefficients of $A(x)$ with the first differences of these coefficients (which are all zero for the geometric series save the zeroth.) Multiplying by further factors of $(1-x)$ gives the second differences, third differences, and so forth.
For the case of $N=1$, we note that the sequence $\{0,1,2,3,\ldots\}$ has first differences $\{0,1,1,1,\cdots\}$ and second differences $\{0,1,0,0,\ldots\}$. Consequently $S_1(x)=\dfrac{x}{(1-x)^2}$. For $N=2$, we need to go up to third differences:
$$\{0,1,4,9,16,\ldots\}\to \{0,1,3,5,7,\ldots\}\to \{0,1,2,2,2,\ldots\}\to \{0,1,1,0,0,\ldots\}$$
and from this we can read off the form of $S_2(x)$ directly (what is it?). The obvious pattern here is that the $(N+1)$-th differences of $\{i^n\}$ have finitely many nonzero entries, and so $(1-x)^{N+1}S_N(x)$ is some polynomial in $x$. (Actually finding this polynomial isn't so nice since it requires computing up to the $(N+1)$ difference, so this really only works well if $a_n$ is a polynomial of low degree in $n$.)
