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 $(a\_1,a\_2,\dots,a\_n)$.



Matrix obtained by recursive multiplication and a cyclic permutation
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](https://en.wikipedia.org/wiki/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](https://en.wikipedia.org/wiki/Circulant_matrix) $(1, \ldots, p'\_{n - 1})$. 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.


