Bernoulli's Inequality/Proof 2

Proof
Let $y = 1 + x$.

Then $y \ge 0$, and:
 * $\paren {1 + x}^n = 1 + \paren {y^n - 1}$

If $y \ge 1$, then by Sum of Geometric Sequence:
 * $\ds y^n - 1 = \paren {y - 1} \sum_{k \mathop = 0}^{n - 1} y^k \ge n \paren {y - 1} = n x$

If $y < 1$, then by Sum of Geometric Sequence:
 * $\ds y^n - 1 = -\paren {1 - y} \sum_{k \mathop = 0}^{n - 1} y^k \ge -n \paren {1 - y} = n x$

Hence the result.