Product of Indices of Real Number/Integers

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

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

Then:


 * $\left({r^n}\right)^m = r^{n m}$

Proof
Proof by induction on $m$:

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

$P \left({0}\right)$ is true, as this just says:
 * $\left({r^n}\right)^0 = 1 = r^0 = r^{n \times 0}$

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

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:
 * $\forall n \in \Z: \left({r^n}\right)^k = r^{n k}$

Then we need to show:
 * $\forall n \in \Z: \left({r^n}\right)^{\left({k + 1}\right)} = r^{n \left({k + 1}\right)}$

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.

Therefore:
 * $(1): \quad \forall n \in \Z, m \in \Z_{\ge 0}: \left({r^n}\right)^m = r^{n m}$

Negative Index
It remains to be shown that:


 * $\forall m < 0: \forall n \in \Z: \left({r^n}\right)^m = r^{n m}$

As $m < 0$ we have that $m = -p$ for some $p \in \Z_{> 0}$.

Thus:

Hence the result, by replacing $-p$ with $m$.