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.

Expand with respect to the first line: the term obtained with $a$ is $aD_{n-1}$. For the second one, we get $(-1)^{n+1}$ times a determinant that can be expanded with respect to the first column. This lead to the recurrence relation 
$$
D_n=aD_{n-1}-\left(n-1\right)^2a^{n-2}.
$$
Letting $b_n:=a^{-n}D_n$ for $a\neq 0$ allows to derive and easier recurrence relation whose resolution shows that the formula mentioned in the opening post, namely, 
$$
D_n=a^n - a^{n-2}\sum_{i=1}^{n-1}i^2,
$$
is correct.
