You often see this I think, in "pretty" proofs compared with the more direct approach. A clever early step or some bit of startling insight.
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?
Look at two consecutive terms: n^(1/n) and (n+1)^[1/(n+1)]. The ratio of the first to the second -- we want to prove that this is >1 -- is n^(1/n) / (n+1)^[1/(n+1)]. This is bigger than 1 iff its n(n+1)th power is; that is, iff n^(n+1) / (n+1)^n > 1. We can write this as n (n/(n+1))^n; so what we need is that (n/(n+1))^n > 1/n.
(Handwavily: we know that the LHS is about 1/e, so for n>=3 this should be good. But we want a proof and we're trying to do it without nontrivial analytic machinery.)
Actually, I prefer to write this as ((n+1)/n)^n < n, or (1+1/n)^n < n. Expand this with the binomial theorem: we get sum {0<=k<=n} of (n choose k) n^-k. And we have (n choose k) = n(n-1)...(n-k+1) / k! < n^k/k! so this is strictly less than sum {0<=k<=n} of 1/k!. And for k>0 we easily have k! >= 2^(k-1) so this sum is no bigger than 1 + 1 + 1/2 + 1/4 + 1/8 + etc. = 3.
So, the function on integers n -> n^1/n is decreasing for n >= 3. Now the proof goes through as before.
(Maybe there's a more thoroughly number-theory-ish way to do it by looking at prime factorizations, but when I try it that way it always seems to end up rather a mess.)
[EDITED to add:] But elsewhere in the discussion users bustermellotron and diffeomorphism give (very similar) neat number-theory-ish proofs, either of which is definitely a better proof than the one using the calculations above.
I suppose at least some proofs vaguely like that are possible because Gödel's incompleteness theorem is one, although I suspect that that same theorem puts some constraints on these "metaproofs."
Inference https://en.wikipedia.org/wiki/Inference ; inductive, deductive, abductive
Propositional calculus > Proofs in propositional calculus: https://en.wikipedia.org/wiki/Propositional_calculus
Quantum logic: https://en.wikipedia.org/wiki/Quantum_logic
https://twitter.com/westurner/status/1609495237738496000 :
Is quantum logic the correct logic for propositional logic?
What are quantum "expectation values"; and how is that Axiomatic wave operator system different from standard propositional calculus?
Yep, I took several metamathematics classes at UCLA! Mostly on proof theory, but also analyzing metamathematical results (like Godel's theorems, Henkin construction, and so on).
I think it's arguable that OP's approach is cleaner for some value of clean, but the article's approach gave me happy flashbacks to doing number theory in undergrad that OP's approach didn't.
When communicating to the wider world, aesthetics do matter, and yeah, everything you said as well.
Differentiating the above function yields (1/x^2)(1-log(x))x^(1/x), which is positive when log(x) < 1 and negative when log(x) > 1. So the function has a maximum at e and decreases on either side of it. Therefore one of our integers must be less than e, and the other greater than it. For the smaller integer there are only two possibilities to check, 1 and 2. Using 1 doesn't give a solution since the equation x^(1/x) = 1 only has the solution 1. So the only remaining possibility for the smaller number is 2, which does yield the solution 2^4 = 4^2. Since x^(1/x) is strictly decreasing when x > e, there can't be any other solutions with the same value.