Okay, given A^B= B^A
Let's define that A >= B, so if they're ever different, A is the bigger one.
Let's define a constant K such that A = KB, then we'll try to prove that K must equal 1.
A^B = B^A
We'll substitute our identity in the right side exponent
A^B = B^(KB)
Well take the Bth root of both sides
A = B^K
We'll substitute our identity in the left side
KB = B^K
and divide by B
K = B^(K-1)
Hmmm. I don't see a way to prove that impossible.
So, let's substitute simple answers here and see if we're getting anywhere-
if K = 2, then what are B and A?
2 = B^(2-1) = B = 2
A = K B = 2*2 = 4
4^2 = 2^4 ? 16 = 16 check
THEORY DISPROVEN!.
if K = 3, then what are B and A?
3 = B^(3-1) = B^2 then B = sqrt(3)
A = KB = 3 sqrt(3)
(3 sqrt(3))^(sqrt(3)) = sqrt(3) ^ (3sqrt(3)) ?
here I resort to a calculator, 17.362 = 17.362 check.
THEORY DISPROVEN!.
There's a whole family of counterexamples.
By the way, you don't need to do any Ks under 1, because you'll just get the same pair in a different order. For instance, k = 1/2 will get you the pair 2,4 again.