| Here goes. 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? |