We can simply calculate the determinant of an opposite (lower) triangular matrix:
Let $J_n$ be the $n \times n$ matrix with $1$ on the anti-diagonal and $0$ otherwise (i.e. $J_ne_i = e_{n+1-i}$ for all $1 \leq i \leq n$, where $e_1, \dotsc, e_n$ denotes the standard basis). Given any $m \times n$-matrix $A$ the matrix $AJ_n$ originates from $A$ by vertically mirroring its colums from the middle, i.e. swapping the first column with the last, the second with the second last, etc.
If $A$ is an $n \times n$ square matrix then we get from $J_n^2 = I_n$ that
$$
\det(A) = \det(J_n) \det(AJ_n).
$$
In the case of $D_n$ we get that $D_n J_n$ is the $a_n$-scalar multiple a lower triangular matrix with diagonal entries $x-a_1, x-a_2, \dotsc, x-a_{n-1}, 1$, so
$$
\det(D_n)
= \det(J_n) \det(D_n J_n)
= \det(J_n) a_n (x-a_1) \dotsm (x-a_{n-1}).
$$
So the only difference is that we need to know $\det(J_n)$. Because $J_n$ is a permutation matrix, corresponding to $\sigma_n \in S_n$ with $\sigma(i) = n+1-i$, we have $\det(J_n) = \mathrm{sgn}(\sigma_n)$. Notice that
\begin{align*}
\sigma_{2n} &= (1 \;\; 2n) (2 \;\; n-1) \dotsm (n \;\; n+1) \\
\sigma_{2n+1} &= (1 \;\; 2n+1) (2 \;\; n-1) \dotsm (n \;\; n+2).
\end{align*}
So we can just count the number of transpositions used and get that
$$
\mathrm{sgn}(\sigma_n) =
\begin{cases}
\phantom{-}1 & \text{if $n \equiv 0,1 \bmod 4$}, \\
-1 & \text{if $n \equiv 2,3 \bmod 4$},
\end{cases}
= (-1)^{n(n-1)/2}.
$$
So alltogether we have
$$
\det(D_n) = (-1)^{n(n-1)/2} a_n (x-a_1) \dotsm (x-a_{n-1}).
$$
(The nice thing about this is that now that we have calculated $\det(J_n) = (-1)^{n(n-1)/2}$ we can use this to calculate the determinant of opposite triangular and opposite diagonal matrices more ore less in the usual way.)
Proof without the use of eigenvalues :
Let $$A=\begin{pmatrix}a_1&a_2&a_3&\cdots&a_n\\a_n&a_1&a_2&\cdots&a_{n-1}\\a_{n-1}&a_n&a_1&\cdots&a_{n-2}\\\vdots&\vdots&\vdots&&\vdots\\a_2&a_3&a_4&\cdots&a_1\end{pmatrix}$$
and $\Omega$ be the matrix with entries $\Omega= (\omega^{(i-1)(j-1)})_{1 \leq i,j \leq n} \in \mathcal{M}_n(\mathbb{C})$, where $\omega = e^{2i\pi/n}$.
Then it is easy to see that the $(i,j)$-entry of the product $A\Omega$ is $\omega^{(i-1)(j-1)}f(\omega^{j-1})$, where $f(x)=a_1+a_2x+a_3x^2+\cdots+a_nx^{n-1}$.
Hence $$A\Omega=\begin{pmatrix}f(1)&f(\omega)&\cdots&f(\omega^{n-1})\\f(1)&\omega f(\omega)&\cdots&\omega^{n-1}f(\omega^{n-1})\\\vdots&\vdots&&\vdots\\f(1)&\omega^{n-1}f(\omega)&\cdots&\omega^{(n-1)(n-1)}f(\omega^{n-1})\end{pmatrix}$$
so
$$\det(A\Omega)=f(1)f(\omega)\cdots f(\omega^{n-1})\begin{vmatrix}1&1&\cdots&1\\1&\omega &\cdots&\omega^{n-1}\\\vdots&\vdots&&\vdots\\1&\omega^{n-1}&\cdots&\omega^{(n-1)(n-1)}\end{vmatrix} $$ $$=\det(A)\det(\Omega)=f(1)f(\omega)\cdots f(\omega^{n-1})\det(\Omega).$$
Because $\det(\Omega) \neq 0$ (it is a Vandermonde), you deduce that
$$\boxed{\det(A)=f(1)f(\omega)\cdots f(\omega^{n-1})}.$$
Best Answer
By linearity in or Laplace expansion along the first column, the determinant is equal to $$ \begin{vmatrix} -1&1&\cdots & 1 \\ -a_1&a_2&\cdots &a_n\\ -a_1^{2}&a_2^{2}&\cdots&a_n^{2}\\ \vdots&\vdots&\ddots& \vdots\\ -a_1^{n-1}&a_2^{n-1}&\cdots&a_n^{n-1}\\ \end{vmatrix} + 2\begin{vmatrix} a_2&\cdots &a_n\\ a_2^{2}&\cdots&a_n^{2}\\ \vdots&\ddots& \vdots\\ a_2^{n-1}&\cdots&a_n^{n-1}\\ \end{vmatrix}. $$ Now each of the two summands is a multiple of a Vandermonde determinant.