Power Function is Strictly Increasing over Positive Reals/Natural Exponent

Theorem
Let $n \in \N$.

Let $f: \R_{> 0} \to \R$ be the real function defined by $f \left({ x }\right) = x^n$.

Then $f$ is strictly increasing.

Proof
For all $n \in \N$, let $P \left({n}\right)$ be the proposition:
 * $0 < x < y \implies 0 < x^n < y^n$

Basis for the Induction
$P(1)$ is true, since $0 < x < 1 \implies 0 < x^1 < y^1$ by definition of exponent of $1$.

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 1$, then it logically follows that $P \left({k+1}\right)$ is true.

So this is our induction hypothesis:


 * $0 < x < y \implies 0 < x^k < y^k$

Then we need to show:


 * $0 < x < y \implies 0 < x^{k+1} < y^{k+1}$

Induction Step
This is our induction step:

First:

Further:

So:

Thus:

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

Therefore:
 * $0 < x < y \implies 0 < x^n < y^n$

Hence the result.