Bernoulli's Inequality/Proof 1

Theorem
Let $x \in \R$ be a real number such that $x \ne -1$.

Let $n \in \Z_{\ge 0}$ be a positive integer.

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

Proof
Proof by induction:

For all $n \in \Z_{\ge 0}$, let $P \left({n}\right)$ be the proposition:
 * $\left({1 + x}\right)^n \ge 1 + nx$

Basis for the Induction
$P \left({0}\right)$ is the case:
 * $\left({1 + x}\right)^0 \ge 1$

so $P \left({0}\right)$ holds.

This is our basis for the induction.

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

So this is our induction hypothesis:
 * $\left({1 + x}\right)^k \ge 1 + kx$

We need to show that:
 * $\left({1 + x}\right)^{k+1} \ge 1 + \left({k + 1}\right) x$

Induction Step
This is our induction step:

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