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.