Logs will appear no matter what, at some point in the line of reasoning. They are only held back until differentiating in this approach. I think the expression for the derivative of logn/n was much nicer to grapple with.
Seeing as we're in the integer world, I'm not sure logs need to show up ever, there's probably a proof path around unique prime factorisations. E.g. it's more or less immediately obvious that n and m would need to share the same prime factors in the same ratios.
The key thing you need is that for n >= 3 the sequence of (n^1/n) is decreasing. You can get that pretty easily without logs.
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.
This back-and-forth makes me wonder, is it possible to write proofs about proofs? e.g., you cannot prove this result without logarithms or there must exist a proof that involves prime factors?
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."
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).
Agreed! I'm realizing that my comment came across as negative, but I did appreciate seeing a second path to the same place. I also agree that the expression x^(1/x) feels like a more natural place to start.
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.