Powers of Coprime Numbers are Coprime

Theorem
Let $a, b$ be coprime integers:
 * $a \perp b$

Then:
 * $\forall n \in \N_{>0}: a^n \perp b^n$


 * If two numbers be prime to one another, and each by multiplying itself make a certain number, the products will be prime to one another; and if the original numbers by multiplying the products make certain numbers, the latter will also be prime to one another [and this is always the case with the extremes].

Proof
Proof by induction:

Let $a \perp b$.

For all $n \in \N_{> 0}$, let $P \left({n}\right)$ be the proposition:
 * $a^n \perp b^n$

$P \left({1}\right)$ is true, as this just says:
 * $a \perp b$

Basis for the Induction
By :
 * $a^2 \perp b$

Again, by :
 * $a^2 \perp b^2$

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k+1}\right)$ is true.

So this is our induction hypothesis:
 * $a^k \perp b^k$

Then we need to show:
 * $a^{k+1} \perp b^{k+1}$

Induction 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:
 * $\forall n \in \N_{>0}: a^n \perp b^n$