Natural Number Multiplication is Commutative/Proof 2

Proof
Proof by induction:

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

For all $n \in \N$, let $\map P n$ be the proposition:
 * $\forall m \in \N: m \times n = n \times m$

Basis for the Induction
From Zero is Zero Element for Natural Number Multiplication:
 * $\forall m \in \N: m \times 0 = 0 = 0 \times m$

Thus $\map P 0$ is seen to be true.

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $\map P k$ is true, where $k \ge 0$, then it logically follows that $\map P {k^+}$ is true.

So this is our induction hypothesis $\map P k$:
 * $\forall m \in \N: m \times k = k \times m$

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

Induction Step
This is our induction step:

So $\map P k \implies \map P {k^+}$ and the result follows by the Principle of Mathematical Induction.

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