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*}
$x^{n-1}a^{0}+x^{n-2}a^{1}+...+x^{1}a^{n-2}+x^{0}a^{n-1}=x^{n-1}\big((\frac{a}{x})^{0}+(\frac{a}{x})^{1}+...+(\frac{a}{x})^{n-2}+(\frac{a}{x})^{n-1}\big)$
$=x^{n-1}\frac{1-(\frac{a}{x})^{n}}{1-\frac{a}{x}}$
