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.

You can just go for calculating the characteristic polynomial $\chi_{-A}$ of minus the matrix $A$ (the one at $a=0$), then your determinant will be $\chi_{-A}[a]$. As you already found that the rank of $A$ is$~2$ (if $n\geq2$; otherwise it is $0$) the coefficients of $\chi_{-A}$ in all degrees less than $n-2$ are zero (as its coefficient of degree $n-r$ is the sum of all principal $r$-minors of$~A$. Since $A$ has zero trace, one has $$\chi_{-A}=X^n+0x^{n-1}+c_nX^{n-1}$$ where $c_n$ is the sum of all principal $2$-minors of$~A$, which is easily seen to be
$$ c_n=-\sum_{k=0}^{n-1}k^2=-\frac{2n^3-3n^2+n}6.$$
Therefore $\det(D_n)=a^n+c_na^{n-2}=a^{n-2}(a^2+c_n)$, as you guessed (with $c_n\leq0$, so the roots of $\chi_{-A}$ are real: $\pm\sqrt{-c_n}$ and $0$ with multiplicity $n-2$).
