This isn't that rigorous, in that assumes that axioms about the addition of natural numbers have already been shown and proven, and I think it also assumes distributivity of multiplication. I don't know if the distributive property is something that has to be shown or assumed.
Anyway, for any natural number $n$, we have that $$n*1 = 1*n $$ because the left hand side is $$ \underbrace{1+ 1 + \dots + 1}_{n\ \text{times}} $$ which is the definition of $n$. Likewise, the right hand side is defined to be $n$. So they are equal.
Now we can use this to show that multiplication with $2$ is commutative, i.e. for any $n$ $$n * 2 = 2 *n$$ The left hand side equals $$ n * (1+1) = n*1 + n*1 = n + n = 1*n + 1*n = (1+1)*n = 2*n$$ Then we can show that multiplication with $3$ is commutative knowing that multiplication with both $1$ and $2$ is commutative, and that multiplication with arbitrary $m$ is commutative knowing that multiplication with $1$ and $m-1$ is commutative.
Is this correct? I see at least two potential problems:
It assumes the distributive property of multiplication for natural numbers. While I find this axiom to be more "common sense" than commutativity, perhaps it should also be proved somehow. And maybe the proof relies essentially on commutativity of multiplication, leading to circular reasoning.
It seems to use not only regular induction, but strong induction. In other words, I am not sure if multiplication being commutative for $m-1$ can be used as the induction hypothesis in regular induction or not, since it seems like it uses implicitly that multiplication is commutative for all lower natural numbers. I think ultimately this comes down to my not understanding the difference between induction and strong induction -- they are supposed to be equivalent, but I imagine that at such a foundational level the distinction between them might be important. What's the difference between simple induction and strong induction?
Context:
This might be related to this question: Why Does Induction Prove Multiplication is Commutative? To be honest I am not sure though since I don't understand that question.
I remember reading somewhere, after doing a Google search, that one can prove that multiplication of natural numbers is commutative by induction. I didn't understand the argument at the time, probably because I didn't put in the effort to understand it properly, but I think I have the idea now, and want to confirm or verify that this is correct. It was stated similar to this: proof of commutativity of multiplication for natural numbers using Peano's axiom
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