Product of Indices of Real Number/Positive Integers

Theorem
Let $r \in \R_{> 0}$ be a positive real number. Let $n, m \in \Z_{\ge 0}$ be positive integers.

Let $r^n$ be defined as $r$ to the power of $n$.

Then:


 * $\paren {r^n}^m = r^{n m}$

Proof
Proof by induction on $m$:

For all $m \in \Z_{\ge 0}$, let $\map P m$ be the proposition:
 * $\forall n \in \Z_{\ge 0}: \paren {r^n}^m = r^{n m}$

$\map P 0$ is true, as this just says:
 * $\paren {r^n}^0 = 1 = r^0 = r^{n \times 0}$

Basis for the Induction
$\map P 1$ is true, by definition of power to an integer:
 * $\paren {r^n}^1 = r^n = r^{n \times 1}$

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:
 * $\forall n \in \Z_{\ge 0}: \paren {r^n}^k = r^{n k}$

Then we need to show:
 * $\forall n \in \Z_{\ge 0}: \paren {r^n}^{\paren {k + 1} } = r^{n \paren {k + 1} }$

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.

Therefore:
 * $\forall n, m \in \Z_{\ge 0}: \paren {r^n}^m = r^{n m}$