Determinant of nth order I want to solve the following determinant:
$D_n= \begin{vmatrix}
        a_n & a_{n-1} & \cdots & a_2 & x\\
        a_n & a_{n-1} & \cdots & x & a_1\\
        a_n & a_{n-1} & \cdots & a_2 & a_1\\
        \vdots & \vdots & \ddots & \vdots & \vdots\\
        a_n & x & \cdots & a_2 & a_1\\
        a_n & a_{n-1} & \cdots & a_2 & a_1\\
        \end{vmatrix}$
The idea I had was to 1) get the $a_n$ in front of the determinant, which gets me to:
$D_n= a_n \begin{vmatrix}
        1 & a_{n-1} & \cdots & a_2 & x\\
        1 & a_{n-1} & \cdots & x & a_1\\
        1 & a_{n-1} & \cdots & a_2 & a_1\\
        \vdots & \vdots & \ddots & \vdots & \vdots\\
        1 & x & \cdots & a_2 & a_1\\
        1 & a_{n-1} & \cdots & a_2 & a_1\\
        \end{vmatrix}$
Then I multiplied the nth row with $-1$ and added it to all the other ones, which gives me:
$D_n= a_n \begin{vmatrix}
        0 & 0 & \cdots & 0 & x-a_1\\
        0 & 0 & \cdots & x-a_2 & 0\\
        0 & 0 & \cdots & 0 & 0\\
        \vdots & \vdots & \ddots & \vdots & \vdots\\
        0 & x-a_{n-1} & \cdots & 0 & 0\\
        1 & a_{n-1} & \cdots & a_2 & a_1\\
        \end{vmatrix}$
So, I got a triangular determinant, but with the opposite diagonal, not the main one. How do I transform it into a real triangular determinant? Obviously, my idea to replace every adjacent row (1st and nth, (n-1)th and 2nd...) is wrong?
Any ideas? 
 A: 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.)
A: You have gone almost through most of the way! The determinant then simply is:$$D=(-1)^{n+1}(x-a_1)(-1)^{n+1}(x-a_2)(-1)^{n+1}(x-a_3)\cdots (-1)^{n+1}(x-a_{n-1})(-1)^{n+1}=(x-a_1)(x-a_2)\cdots (x-a_{n-1})$$
