Why can multiplication be done in any order? I have proof of $ab=ba$ and $abc=acb=bac=cab=cba=bca$ when the terms are all positive. But what about when they are not positive? What about when any number of terms multiplied together? I can't understand why we can rearrange the terms in any way possible. In case of two numbers, I can think of it as area and in case of three I can think of it as volume. But when it exceeds past that it gets difficult. And area and volume only work for positive numbers. What about when the numbers are negative?
 A: The body of your question appeals to geometric intuition about area and volume. The commutativity and associativity of multiplication has an interpretation in this intuitive setting. Namely, that the x, y, z axes do not have an inherent order.
When you build something like a cube or a square and compute its volume, you are imposing an order on the axes, but that choice was arbitrary.
That's one way of thinking about it.
Also, in linear algebra, you do consider signed volume (i.e. the determinant) and in this setting permuting the axes flips signs exactly when the permutation is composed of an odd number of swaps.

For real numbers, the proof of commutativity or associativity is a direct consequence of the real axioms. Alternatively, you can prove it for real numbers using concrete constructions like Dedekind cuts or Cauchy sequences. Basically there are two families of approaches to defining what a real number is. You can build a real number out of simpler concrete things ... or you can specify up front what rules the real numbers must follow.
With that out of the way, proving commutativity and associativity of multiplication for natural numbers is pretty straightforward.
As an example, let's prove commutativity of multiplication for natural numbers.
We will proceed using induction.
First, for zero, $0*a = a*0$.
Let we want to show that, if $k*a = a*k$, then $(k+1)*a = a*(k+1)$.
First, by assumption $k*a = a*k$.
Thus, $k*a + a = a * k + a$ .
Thus, $k*a + 1*a = a*k + a*1$ .
Applying distributivity, we get $(k+1)*a = a * (k+1)$.
