Power of Strictly Positive Real Number is Strictly Positive/Positive Integer

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

Let $n \in \Z_{\ge 0}$ be a positive integer.

Then:
 * $x^n > 0$

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

Proof
Proof by Mathematical Induction:

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

$P \left({0}\right)$ is true, as this just says:

Basis for the Induction
$P \left({1}\right)$ true, as this just says:

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:
 * $\forall x \in \R_{>0}: x^k > 0$

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

Inductive Step
This is our induction step:

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

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