Product of differences of circular permutation Numbers $1,2,\ldots,n$ are arranged into a circle. What is the maximum product of the differences $|x_1-x_2|\times|x_2-x_3|\times\cdots\times|x_{n-1}-x_n|\times|x_n-x_1|$?
I think the maximum should occur when the numbers are arranged $n,1,n-1,2,n-2,3,\ldots$. The sum for this arrangement is $(n-1)(n-2)\cdots1\cdot\lfloor n/2\rfloor = (n-1)!\cdot\lfloor n/2\rfloor$.
This question was inspired by the question asking for the maximum sum. There, it is possible to prove optimality by noting that we have $2n$ terms ($n$ with $+$ and $n$ with $-$), and each number occurs twice.
Here, it is still true that we have $2n$ terms ($n$ with $+$ and $n$ with $-$), and each number occurs twice. But since we're taking the product instead of the sum, optimality is no longer clear.

Your conjecture is false.  Just generating some random permutations, for example, I find a permutation of 10, $2, 6, 1, 7, 3, 9, 5, 10, 4, 8$, for which the product of differences is $8294400$; compare $9! \times 5 = 1814400$ for yours.  Note that all the differences in that permutation are 4, 5, or 6 (that is, near 10/2); I think the small factors in your product towards the end hurt you more than the large ones help.
As a result it makes sense to have as many factors near $n/2$ as possible, and I think something like $1, (n/2+1), 2, (n/2)+2, \cdots, n/2, n$ (for $n$ even) will do quite well - in this case you get a product of $(n/2)^{n/2} (n/2-1)^{n/2-1} (n-1)$.  
