De Moivre's Formula/Positive Integer Index/Proof 1

Theorem
Let $z \in \C$ be a complex number expressed in complex form:
 * $z = r \left({\cos x + i \sin x}\right)$

Then:
 * $\forall n \in \Z_{> 0}: \left({r \left({\cos x + i \sin x}\right)}\right)^n = r^n \left({\cos \left({n x}\right) + i \sin \left({n x}\right)}\right)$

Proof
Proof by induction:

For all $n \in \Z_{> 0}$, let $P \left({n}\right)$ be the proposition:
 * $\left({r \left({\cos x + i \sin x}\right)}\right)^n = r^n \left({\cos \left({n x}\right) + i \sin \left({n x}\right)}\right)$

$P \left({1}\right)$ is the case:


 * $\left({r \left({\cos x + i \sin x}\right)}\right)^1 = r^1 \left({\cos \left({1 x}\right) + i \sin \left({1 x}\right)}\right)$

which is trivially true.

Basis for the Induction
$P \left({2}\right)$ is the case:


 * $\left({r \left({\cos x + i \sin x}\right)}\right)^2 = r^2 \left({\cos \left({n x}\right) + i \sin \left({2 x}\right)}\right)$

From Product of Complex Numbers in Polar Form, we have:
 * $r_1 \left({\cos x_1 + i \sin x_1 }\right) r_2 \left({\cos x_2 + i \sin x_2}\right) = r_1 r_2 \left({\cos \left({x_1 + x_2}\right) + i \sin \left({x_1 + x_2}\right)}\right)$

Setting $r_1 = r_2 = r$ and $x_1 = x_2 = x$ gives the result.

This is our basis for the induction.

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

So this is our induction hypothesis:
 * $\left({r \left({\cos x + i \sin x}\right)}\right)^k = r^k \left({\cos \left({k x}\right) + i \sin \left({k x}\right)}\right)$

Then we need to show:
 * $\left({r \left({\cos x + i \sin x}\right)}\right)^{k+1} = r^{k+1} \left({\cos \left({\left({k+1}\right) x}\right) + i \sin \left({\left({k+1}\right) x}\right)}\right)$

Induction Step
This is our induction step:

Hence, by induction, for all $n \in \Z_{> 0}$:


 * $\left({r \left({\cos x + i \sin x}\right)}\right)^n = r^n \left({\cos \left({n x}\right) + i \sin \left({n x}\right)}\right)$