Natural Number Multiplication is Commutative/Proof 2

Theorem
The operation of multiplication on the set of natural numbers $\N$ is commutative:
 * $\forall m, n \in \N: m \times n = n \times m$

Proof
Proof by induction:

From the definition of natural number multiplication, we have that:

For all $n \in \N^*$, let $P \left({n}\right)$ be the proposition:
 * $\forall m \in \N: m \times n = n \times m$

Basis for the Induction
From Natural Number Multiplication Commutes with Zero, we have:
 * $\forall m \in \N: m \times 0 = m = 0 \times m$

Thus $P \left({0}\right)$ is seen to be true.

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 $P \left({k^+}\right)$ is true.

So this is our induction hypothesis $P \left({k}\right)$:
 * $\forall m \in \N: m \times k = k \times m$

Then we need to show that $P \left({k^+}\right)$ follows directly from $P \left({k}\right)$:
 * $\forall m \in \N: m \times k^+ = k^+ \times m$

Induction Step
This is our induction step:

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

Therefore:
 * $\forall m, n \in \N: m \times n = n \times m$