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.

When I first read your question, I expected that it must mean that addition would possess some obscure property that multiplication lacks, after all, both the additive structure and multiplicative structure are abelian groups, so you'd expect something like this to just generalize. But after some thinking, I realized that this wasn't the case, and instead that the problem is that we aren't generalizing properly.
For if we define "applying an operator $f$, $n$ times, ie $f^n$" as the recursive procedure 
$
f^n(x) =
\begin{cases}
x & \text{if n = 0} \\
f^1(f^{n - 1}(x)) & \text{otherwise}
\end{cases}
$
Then this definition actually uses addition, so if we'd want to generalize this procedure properly, we'd need to change our definition of "applying an operator $n$ times" as well. And indeed $a^n$ does equal $(a^2)^{n / 2}$, which induces a better generalization of commutativity.
