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
For $n \ge 0$.

Proof by Mathematical Induction:

Basis for the Induction
From the definition of power:
 * $x^0 = 1 > 0$

This is our basis for the induction.

Induction Hypothesis
Let $x^k > 0$.

We will show that $x^{k + 1} > 0$.

Inductive Step
We have $x > 0$ and $x^k > 0$.

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:
 * $x^{k + 1} = x^k x > 0$

Therefore, the case when $n \ge 0$ is proved.

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:
 * $x^n = \frac 1 {x^{-n}} > 0$