Power Function is Strictly Increasing over Positive Reals/Natural Exponent

Theorem
Let $n \in \Z_{>0}$ be a strictly positive integer.

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

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

Then $f$ is strictly increasing.

Proof
Proof by induction on $n$:

Let $x, y \in \R_{>0}$ be strictly positive real numbers.

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

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

So this is our induction hypothesis:


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

Then we need to show:


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

Induction Step
This is our induction step:

First:

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

Therefore:
 * $\forall n \in \Z_{>0}: x < y \implies x^n < y^n$