Integer Power of Positive Real Number is Positive

Theorem
Let $x \in \R_{>0}$ be a (strictly) positive real number.

Let $n \in \Z$ be an integer.

Then:
 * $x^n > 0$

where $x^n$ denotes the $n$th power of $x$.

Case 1: $n \ge 0$
Proof by Mathematical Induction:

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

Basis for the Induction
$P \left({0}\right)$ is the case:
 * $\forall x \in \R_{>0}: x^0 = 1 > 0$

from the definition of integer power.

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 $, then it logically follows that $P \left({k + 1}\right)$ is true.

So this is our induction hypothesis:
 * $\forall x \in \R_{>0}: x^k > 0$

Then we need to show:
 * $\forall x \in \R_{>0}: x^{k + 1} > 0$

Inductive Step
This is our induction step:

We have $x > 0$.

From the induction step:
 * $x^k > 0$

By the definition of power:
 * $x^{k + 1} = x^k x$

Therefore, by Real Number Axioms: $\R O 2$: Usual ordering is compatible with multiplication:
 * $x^{k + 1} = x^k x > 0$

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

Therefore:
 * $\forall n \in \Z_{\ge 0}: \forall x \in \R_{>0}: x^n > 0$

Case 2
When $n < 0$, by Order of Real Numbers is Dual of Order Multiplied by Negative Number:
 * $-n > 0$

Then, using the above result:
 * $x^{-n} > 0$

Therefore, by Reciprocal of Strictly Positive Real Number is Strictly Positive:
 * $x^n = \dfrac 1 {x^{-n} } > 0$