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 lower 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.

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.

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 lower 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 $$\det A = (-1)^{\lfloor(n - 1) / 2\rfloor} \prod_{k = 0}^{n - 1} p_k \cdot \prod_{k = 0}^{n - 1}\left(\sum_{j = 0}^{n - 1} p'_j e^{\frac{2i\pi jk}{n}} \right).$$

One can see that the problem is as general as arbitrary circulant determinant.

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 lower 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.