I.e. using distributivity to prove commutativity of multiplication seems kosher to me.
Now it must again be warned that my understanding of the distinction between "regular" and strong induction is weak (especially since both principles can be used to prove the other, so in some sense they are logically equivalent? not sure that's actually true),
Here is a proof for all non-negative integers.
We are attempting to show that $ab=ba$, where $a$ and $b$ are non-negative integers. Let’s introduce a new equivalence, $b+e=a$ (i.e. $e$ is defined as the difference between $a$ and $b$; note that if $e=0$ then the proof becomes trivial. Also note that we assume $a \ge b$; this does not affect our result as the variables are symmetrical). Now we write:
1) $ab= \sum_{i=i}^a b $
This is nothing more than stating the definition of b multiplied by a, ie b summed a times. We can also write:
2) $ba=b(b+e)$
since $b+e=a$, by our own definition. We now try to show that equation (2) can be rewritten in the form of equation (1). We expand on equation (2) by writing:
3) $b(b+e)=\sum_{i=1}^b (b+e)$
This is very similar to what we did regarding equation (1), ie $b(b+e)$ is just $(b+e)$ summed $b$ times. Using some properties of addition, we can transform the right hand side of (3) to read:
4) $\sum_{i=1}^b (b+e)=\sum_{i=1}^b b+\sum_{i=1}^b e$
Now what we are going to do is assume the very thing we set out to prove! That is usually a big no-no unless you are using induction, which is basically where this is going.
5) Assume: $\sum_{i=1}^b e =\sum_{i=1}^e b$
We now substitute (5) back into (4) and proceed:
6) $\sum_{i=1}^b (b+e) = \sum_{i=1}^b b + \sum_{i=1}^e b$
The right hand side of (6) can be simplified:
7) $\sum_{i=1}^b b + \sum_{i=1}^e b = \sum_{i=1}^{b+e} b$
Finally, we make use of the fact that $b+e=a$:
8) $\sum_{i=1}^{b+e}b = \sum_{i=1}^a b$
Comparing (8) to (1) yields what we set out to prove:
9) $ab=ba$
Note that the proof hinges on equation (5), which is nothing more than a restatement of that which we set out to prove, namely:
5a) $be=eb$
The advantage that we have now (after going through all that work) is that we have reduced the number space of the original problem; $e$ by definition is less than $a$ (in the case where $e$ is equal to $a$, $b$ is identically $0$, and the whole proof becomes trivial). We can continue in this way to reduce the number space of the problem until we eventually get to a base case which can be shown to be trivially true (namely when $e=0$); this is the nature of inductive proof .
I know this isn’t as formal as a textbook proof, but it is a cute little intuitive proof that I had not yet seen presented in such a way on the internet, so I thought I would submit it. I really hope it helps someone!
Best Answer
The first part is a comment rather than an answer, but I don't have enough points to comment.
It is impossible to answer this question mathematically without more information about what your definition of "natural number", "sum" and "multiplication" are. It is closely tied to the question of how you choose to present the foundations of mathematics.
The most straightforward way to define the sum of two numbers $m$ and $n$ is to take two disjoint sets $A$ and $B$ (disjoint means with no common elements), with $m$ and $n$ elements respectively, and to define $m + n$ as the number of elements in the set $A \cup B$. The product $mn$ is the number of elements in the Cartesian product $A \times B$.
I've left out many important points:
What is a natural number?
What is the precise definition of the "number of elements" in a given finite set $A$? A related question is of course what a finite set is.
How do I know that the sets $A \cup B$ and $A \times B$ are finite?
How do I know that, if I replace $A$ and $B$ with other sets having the same numbers of elements, the number of elements in $A \cup B$ (or $A \times B$) won't change?
These are all questions tied to the foundations of mathematics, and a "proof," in order to be entirely valid, would have to be technical. However, so long as the sets $A \cup B$ and $A \times B$ are established to be finite, there can be no question that the "number of elements" of these sets will be a natural number.
You can learn more about these questions by reading about the mathematical concept of "cardinality."