Calculating a determinant. $D_n$=\begin{vmatrix}
a & 0 & 0 & \cdots  &0&0& n-1 \\
0 & a & 0 & \cdots &0&0& n-2\\
0 & 0 & a & \ddots &0&0& n-3 \\
\vdots & \vdots & \ddots & \ddots & \ddots&\vdots&\vdots \\
\vdots & \vdots & \vdots & \ddots & \ddots& 0&2 \\
0 & \cdots & \cdots & \cdots &0&a&1  \\
n-1 & n-2 & n-3 & \cdots & 2 & 1& a\\
\end{vmatrix}
I tried getting the eigenvalues for A =
\begin{vmatrix}
0 & 0 & 0 & \cdots  &0&0& n-1 \\
0 & 0 & 0 & \cdots &0&0& n-2\\
0 & 0 & 0  & \ddots &0&0& n-3 \\
\vdots & \vdots & \ddots & \ddots & \ddots&\vdots&\vdots \\
\vdots & \vdots & \vdots & \ddots & \ddots& 0&2 \\
0 & \cdots & \cdots & \cdots &0&0&1  \\
n-1 & n-2 & n-3 & \cdots & 2 & 1& 0\\
\end{vmatrix}
For $a=0$  , the rank of the matrix is $2$ , hence $\dim(\ker(A)) = n-2 $  
$m(0)>=n-2$
However, I was not able to determine the other eigenvalues.
Testing for different values of n :
for $n=2$ : 
$D_2 = a^2-1$
for $n=3$ :
$D_3 = a^3 -5a$
$D_n$ seems to be equal to $a^n - a^{n-2}\sum_{i=1}^{n-1}i^2$ .
However I'm aware that testing for different values of $n$ is not enough to generalize the formula.
Thanks in advance.

To find other values of $x$ besides $0$ for which
$$
  \det \begin{bmatrix}
x & 0 & 0 & \cdots & n-1 \\
0 & x & 0 & \cdots & n-2 \\
0 & 0 & x & \cdots & n-3 \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
n-1 & n-2 & n-3 & \cdots & x
\end{bmatrix} = 0
$$
consider taking $\frac{n-1}{x}$ times the first row, plus $\frac{n-2}{x}$ times the second row, plus $\frac{n-3}{x}$ times the third, and so on, on the grounds that you will get a row vector which is equal to the last row in its first $n-1$ entries.
Its last entry is going to be $\frac{(n-1)^2 + (n-2)^2 + \dots + 2^2 + 1^2}{x} = \frac{n(n-1)(2n-1)}{6x}$. So if this happens to be equal to $x$, then the last row is a linear combination of the other rows, which means the determinant is $0$.
There are two values of $x$ for which this works: the two square roots of $\frac{n(n-1)(2n-1)}{6}$. This gives us the last two eigenvalues: the whole list is
$$
    a, a, \dots, a, a - \sqrt{\frac{n(n-1)(2n-1)}{6}}, a + \sqrt{\frac{n(n-1)(2n-1)}{6}}
$$
and their product is
$$
   a^{n-2}\left(a^2 - \frac{n(n-1)(2n-1)}{6}\right)
$$
as you conjectured.
