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.

It is maybe a bit easier to use the matrix determinant formula. If you rewrite your matrix as 
$$
M = aI +UV^T
$$
where $UV^T$ is
$$
\begin{bmatrix}
(n-1) &0\\
(n-2) &0\\
\vdots&0\\
1 &0\\
0&1
\end{bmatrix}
\begin{bmatrix}
0&0&\cdots&1\\
(n-1)&(n-2)&\cdots&0
\end{bmatrix}
$$
Then, from 
$$
\operatorname{det}({\mathbf{A}}+{\mathbf  {UV}}^{{\mathrm  {T}}})=\operatorname {det}({\mathbf{I}}+{\mathbf  {V}}^{{\mathrm  {T}}}{\mathbf  {A}}^{{-1}}{\mathbf  {U}})\operatorname {det}({\mathbf  {A}}).
$$
and $\mathbf{A}^{-1}=\frac1aI$, we have only a $2\times 2$ determinant to evaluate:
$$
\det(M) = \left|\begin{matrix}1&\frac{1}{a}\\\displaystyle\frac{1}{a}\sum_{i=1}^{n-1}i^2&1\end{matrix}\right|a^n = (1-\frac{1}{a^2}\sum_{i=1}^{n-1}i^2)a^n
$$
