Why is exponentiation right associative? From Wikipedia:

In order to reflect normal usage, addition, subtraction,
  multiplication, and division operators are usually left-associative
  while an exponentiation operator (if present) is right-associative

For example, we evaluate $2^{2^2}$ as $2^4$ rather than $4^2$. All other operations besides exponentiation, tetration, etc. are inherently left associative. Is there any reason why it's different for exponents?
 A: As André Nicolas writes, it is because $$  (a^b)^c = \underbrace{(a^b) \cdot (a^b) \cdots (a^b)}_{c\text{-times}} = a^{b \cdot c} \text{,}  $$ so there is no need to use iterated exponentiation to represent that idea.
Additionally, consider the evaluation tree for the expression "a^b^c".  It is
"^(a, ^(b,c))" because, by order of operations, we evaluate the exponent first.
Finally, addition and multiplication are commutative and associative, so it does not matter in what order they are applied.  Claiming that subtraction is left associative may be incomplete, depending on how you define subtraction.  If subtraction is not adding the additive inverse, then it does not acquire commutativity and associativity $$\begin{align*}
 a - b - c &= (a-b)-c \neq a-(b-c)  \\
    &= a-(b+c)  \\
    &= a+ (-b) + (-c).
\end{align*}$$  The same thing happens for division: $$\begin{align*}
 a / b / c &= (a/b)/c \neq a/(b/c)  \\
    &= a/(b \times c)  \\
    &= a(b^{-1})(c^{-1}).
\end{align*}$$  (It is, perhaps, instructive to realize I just wrote the same display twice.)  If we define subtraction and division as in the third lines of the two displays, then these operations are just as associative and commutative as addition and multiplication.
Is there any hope of doing the same thing with exponentiation?  The above come in inverse pairs.  Perhaps we should look at logarithms.  But neither $ \log_a b = \log_b a$ nor $a^b = b^a$, so we don't have an associative and commutative member of the pair to write both operations in terms of.  This is a commutative variation, $a^{\ln b} = \mathrm{e}^{(\ln a) (\ln b)} = b^{\ln a}$, but I can't say this is a common operation or that others would recognize it as rapidly as the six operations discussed so far.  (This idea can be generalized to make commutative variants of tetration and higher operations.)
