Closed-form Geometric series of both increasing and decreasing variables? This question comes from the formula 
$$x^n - a^n = (x-a)(x^{n-1}a^0  + x^{n-2}a^1 + .... + x^1a^{n-2} + x^0a^{n-1})$$ 
which can be verified by summing the second factor as a geometric series. My question is, how do you express the second factor as a geometric series in closed form? If the $x$ factors were instead the constant $1$, we might have 
$$s_n =1\cdot a^0 + 1 \cdot a^1 +.... +1\cdot a^{n-1}=\frac{1\cdot (1-a^n)}{1-a}$$
but that's as far as I've gotten.

Observe that the series has common ratio $r = \frac{a}{x}$. Hence:
\begin{align*}
(x-a)(x^{n-1}a^0  + x^{n-2}a^1 + \ldots + x^1a^{n-2} + x^0a^{n-1})
&= (x-a) \cdot \dfrac{x^{n-1}(1-(\frac{a}{x})^n)}{1-\frac{a}{x}} \\
&= (x-a) \cdot \dfrac{x}{x} \cdot \dfrac{x^{n-1}(1-\frac{a^n}{x^n})}{1-\frac{a}{x}} \\
&= (x-a) \cdot \dfrac{x^{n}(1-\frac{a^n}{x^n})}{x-a} \\
&= x^n - a^n
\end{align*}
