n^m = m^n, so n^m and m^n share the same prime factorization. Call it p1^x1 * ... * pk^xk.
n = n^m^(1/m) = p1^(x1/m) * ... * pk^(xk/m)
m = m^n^(1/n) = p1^(x1/n) * ... * pk^(xk/n)
Therefore each xi is divisible by both n and m, so it's divisible by lcm(n, m). Call lcm(n, m) = d. Now define:
z = p1^(x1/d) * ... * pk^(xk/d)
Both n and m are powers of z!
n = z^(d/m)
m = z^(d/n)
Call a=d/m and b=d/n. Then:
n^m = (z^a)^(z^b) = z^(az^b)
m^n = (z^b)^(z^a) = z^(bz^a)
n^m = m^n -> z^(az^b) = z^(bz^a) -> az^b = bz^a
EDIT: continuing proof.
That's as far as I've got. Can someone run with it?
n^m = m^n, so n^m and m^n share the same prime factorization. Call it p1^x1 * ... * pk^xk.
n = n^m^(1/m) = p1^(x1/m) * ... * pk^(xk/m)
m = m^n^(1/n) = p1^(x1/n) * ... * pk^(xk/n)
Therefore each xi is divisible by both n and m, so it's divisible by lcm(n, m). Call lcm(n, m) = d. Now define:
z = p1^(x1/d) * ... * pk^(xk/d)
Both n and m are powers of z!
n = z^(d/m)
m = z^(d/n)
Call a=d/m and b=d/n. Then:
n^m = (z^a)^(z^b) = z^(az^b)
m^n = (z^b)^(z^a) = z^(bz^a)
n^m = m^n -> z^(az^b) = z^(bz^a) -> az^b = bz^a
EDIT: continuing proof.
That's as far as I've got. Can someone run with it?