Bernoulli's Inequality/Corollary/Proof 2

Proof
Proof by induction:

Let $0 < x < 1$.

For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
 * $\paren {1 - x}^n \ge 1 - n x$

Basis for the Induction
$\map P 0$ is the case:
 * $\paren {1 - x}^0 = 1 \ge 1 - 0 x = 1$

so $\map P 0$ holds.

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $\map P k$ is true, where $k \ge 0$, then it logically follows that $\map P {k + 1}$ is true.

So this is our induction hypothesis:
 * $\paren {1 - x}^k \ge 1 - k x$

We need to show that:
 * $\paren {1 - x}^{k + 1} \ge 1 - \paren {k + 1} x$

Induction Step
This is our induction step:

So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.