Bernoulli's Inequality/Proof 2

Theorem
Let $X$ be one of the sets of numbers $\N$, $\Z$, $\Q$, or $\R$.

Let $n \in \N$, $x \in X$, $x \ge -1$.

Then:
 * $\left({1 + x}\right)^n \ge 1 + nx$

Proof
Let $y = 1 + x$.

Then $y \ge 0$, and:
 * $\left({1 + x}\right)^n = 1 + \left({y^n - 1}\right)$

If $y \ge 1$, then by Sum of Geometric Progression:
 * $\displaystyle y^n - 1 = \left({y - 1}\right) \sum_{k \mathop = 0}^{n-1} y^k \ge n \left({y - 1}\right) = nx$

If $y < 1$, then by Sum of Geometric Progression:
 * $\displaystyle y^n - 1 = -\left({1 - y}\right) \sum_{k \mathop = 0}^{n-1} y^k \ge -n \left({1 - y}\right) = nx$

Hence the result.