Natural Number Multiplication is Commutative/Proof 3

Theorem
The operation of multiplication on the set of natural numbers $\N_{> 0}$ is commutative:
 * $\forall x, y \in \N_{> 0}: x \times y = y \times x$

Proof
Using the axiom schema:

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

Basis for the Induction
$P \left({1}\right)$ is the case:

and so $P \left({1}\right)$ holds.

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:
 * $\forall a \in \N: a \times k = k \times a$

Then we need to show:
 * $\forall a \in \N: a \times \left({k + 1}\right) = \left({k + 1}\right) \times a$

Induction Step
This is our induction step:

The result follows by the Principle of Mathematical Induction.