Real Number to Negative Power/Positive Integer

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

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

Then:


 * $r^{-n} = \dfrac 1 {r^n}$

Proof
Proof by induction on $m$:

For all $n \in \Z_{\ge 0}$, let $P \left({n}\right)$ be the proposition:
 * $r^{-n} = \dfrac 1 {r^n}$

$P \left({0}\right)$ is the case:

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

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:
 * $r^{- k} = \dfrac 1 {r^k}$

Then we need to show:
 * $r^{- \left({k + 1}\right)} = \dfrac 1 {r^{k + 1} }$

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:
 * $\forall n \in \Z_{\ge 0}: r^{-n} = \dfrac 1 {r^n}$