commutative power

The power operator ("a raised to the power of b") is certainly not commutative, i.e. in general a^b ≠ b^a. This should be no surprise, e.g. 2³ ≠ 3².

What I am interested is find pairs of natural numbers, in which a^b does equal b^a. I already know one pair: 2⁴ = 4² (in addition to the trivial pairs, e.g. 3³ = 3³), so I need to find out what's so special about this pair of numbers.

To do this, I break the equation into a series of steps:
2⁴ = 2^2×2 = (2²)² = 4².
Now the relationship becomes clear; I want a pair of numbers such that:
a^b = a^c×a = (a^c)^a = b^a
where c is another natural number that satisfies:
b = c×a and b = a^c
I can combine these equations into one:
c a = a^c
So how many (nontrivial) pairs of natural numbers satisfy this equation? By nontrivial, I exclude cases where c = 1.

Now, to solve this equation:
c a = a^c
c a - a^c = 0
(c - a^{c - 1}) a = 0
Either a = 0 (which is trivial) or c - a^{c - 1} = 0. Consider the latter nontrivial case:
c - a^{c - 1} = 0
c = a^{c - 1}
c^{1/(c - 1)} = a
Therefore, if a value of c is known, a can be found by taking the (c - 1)^th root of c. If c = 2, then a = 2^{1/(2 - 1)} = 2, and therefore b = c a = 4; this shows why 2⁴ = 4². For any other c > 2, (c - 1)^th root of c will not be an integer, so there are no other nontrivial natural-number combinations of a, b and c.

Well, looks like that answered the question. It seems 2⁴ = 4² is pretty special case, after all.