Why does $a\times b=b\times a$ for $a,b \in \mathbb{N}$? Why does $a\times b=b\times a$ for $a,b \in \mathbb{N}$ ?
I always got told that it is the case, and it seems to be, but is there a proof for it. If we start with $a,b \in \mathbb{N}$ then I could try to work out $\mathbb{Z}$ then $\mathbb{Q}$ then $\mathbb{R}$ then $\mathbb{C}$. 
As far as I know $a\times b = \underbrace{b+b+ \ldots +b}_{a\text{ times}}$ .
But why would that necessarily equal $\underbrace{a+a+ \ldots +a}_{b\text{ times}}$ .
I know the way to explain it to 12 year old children is by relating it to areas. But I am not quite satisfied with that explanation.

here $a$ and $b$ are interchangeable because you can always rotate the image by $90°$ while keeping the same area.
This is something my friends and I have had big discussions over. First I tried to figure it out myself, but I didn't get far at all. I also brought it to my teachers and a friend doing a BMath, but all they said is that it is an interesting and strange question. I am not in university yet, so I'm sorry if this is a 'bad' question. 
 A: It's a matter of axioms and definitions and what you think math, multiplication, and natural numbers are.
If we view natural numbers as discrete quantities then counting, adding, and multiplying them are just ways of grouping them.  Our first and most basic assumption is that discrete quantities are and won't change no matter how we group them.
(Here's a brief diversion: Why does $3+4$ always equal $7$?  Why doesn't it sometimes equal $7$ and sometimes equal $8$ and for that matter why does $7$ always equal $7$?  Why can't we turn our back and suddenly see it's turned into $5$ or $8$?  These are possible stupid and meaningless questions.  Or maybe they are profound.  They can't actually be answered except to say: "Because they can't".  At some level we say "It is a fundamental and axiomatic presumption that quantities are constant.")
So as we'd teach children $a \times b$ is the number one has when one has $a$ groups of $b$ items.  (Think: $7$ oranges in $3$ paper pages is $21$ oranges.)  If we manipulate and regroup so that we take one item from each of the $a$ groups, and group them into a new group of $a$ items and do that for each of the $b$ items, we will end of with $b$ groups of $a$ items or $b \times a$.  Now we didn't change any totals.  We just regrouped them.  (Think:  Take one orange from each of the three paper bags and put them in a plastic bag.  $3$ oranges in a plastic bag.  Repeat until you've done it $7$ times for each of the seven oranges in each bag.  You now have $3$ oranges in $7$ plastic bags... and seven empty paper bags.)
By the way.  Why didn't the turning the rectangle $90^\circ$ convince you?  It's clearly the same rectangle with the same tiles but the rows and columns are swapped.
Anyway.  But that is a basic understanding and with any unstated presumption that we are counting things like oranges, or rocks, or tiles.
Being mathematicians, we are never satisfied with such things and are always asking "But what is it really?" And "what justifications do have that we can assume reality is consistent?".  And "who cares about reality anyway? Reality is just a ... physical .... ugh ... application of theory, and surely knowing if the theory is true is much more important and just how the ugh 'real' world just coincidentally behaves."
It which case we have axioms, definitions and constructions.
One approach:  number, addition, and multiplications are just abstract concepts and that $a+b =b+a$ and $a\times c = c\times a$ and the result of $a + b$ and then plus $c$ is the same as the result of $a$ and then adding $b+c$ (or $(a+b) +c = a+(b+c)$) and $a\times(b\times c) = (a\times b)\times c$ and $a(b+c) = (a\times b) + (a\times c)$ are simply given as axioms.  If we are further given that $0 + x = x$ and $1\times x = x$ and that the chain: $0; 0+1=1; 1+1=2; 2+1=3; 3+1 = 4$.... will never repeat we've got the natural numbers.  (Google: Field Axioms and Ordered Field Axioms.)
Another (google Peano Axioms) approach is to systematically define "numbers" and "counting" by considering "successors" ("0" simply exists and nothing is succeded by "0".  1 is the successor of $0$.  All numbers have one and only one successor) and some axioms.  From a long list of axioms and a long succession of definitions (such as $a + b$ means applying the successor operation to $a$, $b$ times, etc.) there is a fundamental (and tedious) proof that $a + b = b+ a$ and $a \times b = b\times a$.
Frankly, I'm happy with the oranges in bags explanation. Or the field axiom.  And I'm glad someone has done the Peano axioms and they are available for me to read but I am grateful I didn't have to figure them out.
A: It's a good question. You must keep in mind that, in order to prove with rigour something about multiplication, there are two things that must be done:


*

*Multiplication must be rigorously defined.

*There must be a set of axioms (about natural numbers) to begin with.


Then we expect to be able to deduce from these axioms that multiplication is commutative.
The most widely used set of axioms for the natural numbers are the Peano axioms. Within this system, you can define multiplication (assuming that the sum of two natural numbers has already been defined) as the only operation such that:


*

*If $n\in\mathbb{N}$, then $n\times1=n$;

*If $n,m\in\mathbb{N}$, then $n\times(m+1)=n\times m+n$.


Using this, you can indeed prove that the multiplication is commutative. I shall do it assuming that addition is commutative (this should be proved too, of course).
Proposition 1: If $n\in\mathbb N$, then $1\times n=n$.
Proof: For $n=1$, this simply asserts that $1\times1=1$, which is true, by the definition of multiplication.
Let $n\in\mathbb N$ and assume that $1\times n=n$. Then\begin{align}1\times(n+1)&=1\times n+1\\&=n+1.\end{align}
Proposition 2: If $m,n\in\mathbb N$, then $(m+1)\times n=m\times n+n$.
Proof: Let $m\in\mathbb N$. If $n=1$, this just asserts that $m+1=m+1$, which is trivially true. Now, let $n\in\mathbb N$ and assume that $(m+1)\times n=m\times n+n$. Then\begin{align}(m+1)\times(n+1)&=(m+1)\times n+m+1\\&=m\times n+n+m+1\\&=m\times n+m+n+1\text{ (addition is commutative)}\\&=m\times(n+1)+n+1\end{align}
Proposition 3: In $\mathbb N$, the multiplication is commutative. In other words, if $m,n\in\mathbb N$, $m\times n=n\times m$.
Proof: If $m\in\mathbb N$, then both $m\times1$ and $1\times m$ are equal to $m$ (because of the definition of multiplication and of the proposition 1).
Now, let $n\in\mathbb N$ and suppose that $m\times n=n\times m$. Then\begin{align}m\times(n+1)&=m\times n+m\\&=n\times m+m\\&=(n+1)\times m,\end{align}by proposition 2.
A: You may apply mathematical induction to prove that for  any two natural numbers $m$ and $n$, $$ mn=nm $$
Fix  $m$  and define $$ P(n) : mn=nm $$
$P(1)$ is true because $m\times 1= 1\times m$
If $P(k)$ is true we need to show that $P(k+1)$ is also true.
That is we assume $mk=km$ and try to prove $m(k+1)=(k+1)m$
Note that $$m(k+1)=mk+m = km+m =(k+1)m$$
That is $P(k+1)$ is true.
A: The level of proof required depends on the context. The explanation with a grid was good enough for Dirichlet's 1863 Zahlentheorie. Here's page $1$. 

The text begins (rough translation - my German is ancient):

In this chapter we treat several facts about arithmetic that (to be
  sure) can be found in most textbooks, but for our work are so
  fundamental that we have to provide a strong foundation.  First, the
  theorem that the product of a list of positive whole numbers is
  independent of the order in which you carry out the multiplications.

He then describes the array, with $a$ copies of $c$ in each of $b$ rows. 
On page $2$ he uses that picture to prove/explain the associativity and commutativity of multiplication.
