Why are addition and multiplication commutative, but not exponentiation? We know that the addition and multiplication operators are both commutative, and the exponentiation operator is not.  My question is why.
As background there are plenty of mathematical schemes that can be used to define these operators. One of these is hyperoperation 
where
$H_0(a,b) = b+1$   (successor op)
$H_1(a,b) = a+b$   (addition op)
$H_2(a,b) = ab $   (multiplication op)
$H_3(a,b) = a^b$   (exponentiation op)
$H_4(a,b) = a\uparrow \uparrow b$   (tetration op: $a^{(a^{(...a)})}$ nested $b$ times )
etc.  
Here it is not obvious to me why $H_1(a,b)=H_1(b,a)$ and $H_2(a,b)=H_2(b,a)$ but not $H_3(a,b)=H_3(b,a)$
Can anyone explain why this symmetry breaks, in a reasonably intuitive fashion?
Thanks.

There's a pretty good heuristic that comes from set theory.  If you have two sets, $X,Y$, then:
$X\times Y=$ all ordered pairs $(x,y):x\in X, y\in Y$
$Y^X=$ all functions $f:X\rightarrow Y$
Now if $X$ and $Y$ happen to be finite sets, with $|X|=m$ and $|Y|=n$, then we have $|X\times Y|=mn$ and $|Y^X|=n^m$ as usual.
So it should be clear that $|X\times Y|=|Y\times X|$, while $|Y^X|\ne|X^Y|$, for finite sets.
Now these rules also happen to go through for infinite sets as well, which takes more work to show.
Finally, there is a quite natural set isomorphism between $X\times Y$ and $Y\times X$, $ (x,y)\leftrightarrow (y,x)$, but certainly not between $Y^X$ and $X^Y$; i.e. you cannot just switch a domain set with its range and expect to have a set isomorphism on the new set of functions (e.g. let one of the sets have just one member and the other set more than one member).
I realize there are many details to fill in here to make this argument fully rigorous, but I think it shows why there is no reason to think that exponentiation should commute given the set-theoretic analogue of functions between sets.
