Matrix obtained by recursive multiplication and a cyclic permutation Have you ever seen this matrix?  Each row is obtained from the previous one by multiplying each element by the corresponding element of the next cyclic permutation of $(a_1,\dots, a_n)$:
$$\left(
  \begin{array}{llllllll}
    1 & 1 & 1 & \dots & 1 & 1 \\
    a_1 & a_2 & a_3 &  \dots & a_{n-1} & a_{n} \\
    a_1 a_2 & a_2a_3 & a_3 a_4 &  \dots & a_{n-1} a_n & a_{n} a_1 \\
    a_1 a_2a_3 & a_2a_3 a_4 & a_3 a_4a_5 &  \dots & a_{n-1} a_n a_1& a_{n} a_1 a_2 \\
   a_1 a_2a_3a_4 & a_2a_3 a_4 a_5& a_3 a_4a_5 a_6& \dots & a_{n-1} a_n a_1 a_2& a_{n} a_1 a_2a_3 \\
    \dots & \dots & \dots & \dots & \dots  & \dots \\
     a_1 a_2a_3a_4 \dots a_{n-2}& a_2a_3 a_4 a_5\dots a_{n-1} & a_3 a_4a_5 a_6\dots a_n&\dots & a_{n-1} a_n a_1 a_2\dots a_{n-4}& a_{n} a_1 a_2a_3\dots a_{n-3} \\
    a_1 a_2a_3a_4 \dots a_{n-1}& a_2a_3 a_4 a_5\dots a_{n} & a_3 a_4a_5 a_6\dots a_1& \dots & a_{n-1} a_n a_1 a_2\dots a_{n-3}& a_{n} a_1 a_2a_3\dots a_{n-2} \\
  \end{array}
\right)$$
I would like to know if there is a closed formula for the determinant; of course it is invariant (up to sign) under cyclic permutations of $(a_1,a_2,\dots,a_n)$.
 A: Denote the matrix $A$, and index all $a_i$, and all rows and columns starting from $0$ for convenience. If, say, $a_0 = 0$, then $\det A = (-1)^{\lfloor (n - 1) / 2 \rfloor} \prod_{i = 1}^{n - 1} a_i^i$ by substituting and computing the remaining upper triangular determinant. Let's further assume that all $a_i$ are non-zero.
Let $p_k = \prod_{i = 0}^{k - 1} a_i$ ($p_0 = 1$ by convention). Put $a'_i = a_i p_n^{-1/n}$, and construct $A'$ similarly. Alternatively, $A'$ is $A$ with $i$-th row multiplied by $p_n^{-i/n}$. Defining $p'_k$ similarly, we now have $p'_k = p_k p_n^{-k / n}$, in particular $p'_n = 1$.
After multiplying $i$-th column of $A'$ by $p'_i$, we arrive at a Hankel matrix $$P' = \begin{pmatrix} 1 & p'_1 & \ldots & p'_{n - 1} \\
p'_1 & p'_2 & \ldots & 1 \\ \ldots & \ldots & \ldots & \ldots \\ p'_{n - 1} & 1 & \ldots & p'_{n - 2}\end{pmatrix},$$ which is a row permutation of a circulant $(1, \ldots, p'_{n - 1})$. With all substitutions in mind, we have
$$\det A = (-1)^{\lfloor(n - 1) / 2\rfloor} \prod_{k = 0}^{n - 1} p_k \cdot \prod_{k = 0}^{n - 1}\left(\sum_{j = 0}^{n - 1} p'_j e^{i\frac{2\pi jk}{n}} \right).$$
One can see that the problem is as general as arbitrary circulant determinant.
