Sum of Geometric Sequence/Proof 3

Proof
From Difference of Two Powers:


 * $\displaystyle a^n - b^n = \left({a - b}\right) \left({a^{n - 1} + a^{n - 2} b + a^{n - 3} b^2 + \ldots + a b^{n - 2} + b^{n - 1} }\right) = \left({a - b}\right) \sum_{j \mathop = 0}^{n - 1} a^{n - j - 1} b^j$

Set $a = x$ and $b = 1$:


 * $\displaystyle x^n - 1 = \left({x - 1}\right) \left({x^{n - 1} + x^{n - 2} + \cdots + x + 1}\right) = \left({x - 1}\right) \sum_{j \mathop = 0}^{n - 1} x^j$

from which the result follows directly.