<s>[INST] Determinant with ones on diagonal and increasing powers of a constant in rows and columns I've found a few determinants of this form and I would like to reach a simple expression for this determinant in terms of a and n.
$$
    D_{a,n}=\begin{vmatrix}
    1 & a & a^2 &\cdots & a^{n-1} & a^n\\
    a & 1 & a &\cdots & a^{n-2} & a^{n-1}\\
    a^2 & a & 1 &\cdots & a^{n-3} & a^{n-2}\\
\vdots & \vdots & \vdots & \ddots &\vdots & \vdots \\
a^{n-1} & a^{n-2} & a^{n-3} & \cdots & 1 & a \\
a^n & a^{n-1} & a^{n-2} & \cdots & a & 1
    \end{vmatrix}_{(n+1) \times (n+1)}
$$
My idea was to subtract the first row multiplied by the leading coefficient (of the other rows) from all others. This will put $ (1-a^2)$ on the main diagonal if repeated n times, while also making $D_{a,n}$ the determinant of an upper triangular matrix.
I.e:
$$
    D_{a,n}=\begin{vmatrix}
    1 & a & a^2 &\cdots & a^{n-1} & a^n\\
    0 & 1-a^2 & a(1-a^2) &\cdots & a^{n-2}(1-a^2) & a^{n-1}(1-a^2)\\
    0 & a(1-a^2) & 1-a^4 &\cdots & a^{n-3}(1-a^4) & a^{n-2}(1-a^4)\\
\vdots & \vdots & \vdots & \ddots &\vdots & \vdots \\
0 & a^{n-2}(1-a^2) & a^{n-3}(1-a^2) & \cdots & 1-a^{2(n-1)} & a(1-a^{2(n-1)}) \\
0 & a^{n-1}(1-a^2) & a^{n-2}(1-a^2) & \cdots & a(1-a^{2(n-1)}) & 1-a^{2n}
    \end{vmatrix}_{(n+1) \times (n+1)}
\\
$$
$$
    D_{a,n}=\begin{vmatrix}
    1 & a & a^2 &\cdots & a^{n-1} & a^n\\
    0 & 1-a^2 & a(1-a^2) &\cdots & a^{n-2}(1-a^2) & a^{n-1}(1-a^2)\\
    0 & 0 & 1-a^2 &\cdots & a^{n-3}(1-a^4) & a^{n-2}(1-a^4)\\
\vdots & \vdots & \vdots & \ddots &\vdots & \vdots \\
0 & 0 & a^{n-3}(1-a^2)^2 & \cdots & 1-a^{2(n-2)} & a(1-a^{2(n-2)}) \\
0 & 0 & a^{n-2}(1-a^2) & \cdots & a(1-a^{2(n-2)}) & 1-a^{2(n-1)}
    \end{vmatrix}_{(n+1) \times (n+1)}
\\ \vdots
$$
$$
D_{a,n}=\begin{vmatrix}
    1 & a & a^2 &\cdots & a^{n-1} & a^n\\
    0 & 1-a^2 & a(1-a^2) &\cdots & a^{n-2}(1-a^2) & a^{n-1}(1-a^2)\\
    0 & 0 & 1-a^2 &\cdots & a^{n-3}(1-a^4) & a^{n-2}(1-a^4)\\
\vdots & \vdots & \vdots & \ddots &\vdots & \vdots \\
0 & 0 & 0 & \cdots & 1-a^2 & something \\
0 & 0 & 0 & \cdots & 0 & 1-a^2
    \end{vmatrix}_{(n+1) \times (n+1)}
$$
$$
\implies D_{a,n}= (1-a^2)^{n}
$$
However, I don't believe this is rigorous enough proof to be used on an exam.
I'm hoping someone could point me in the right direction.
(I've thought of using induction, but it doesn't feel like the simplest approach)
(Also, what would "something" be if done the way I started?)
Edit: I just found out this is the determinant of a Toeplitz matrix, which has already been answered before here.
 [/INST] Denote by $M$ the matrix in question (note that there are $n+1$ entries on the first row of $M$; thus $M$ is $(n+1)\times(n+1)$, not $n\times n$). Then
$$
\pmatrix{1\\ -a&1\\ &\ddots&\ddots\\ &&-a&1}M=\pmatrix{1&\ast&\cdots&\ast\\ &1-a^2&\ddots&\vdots\\ &&\ddots&\ast\\ &&&1-a^2}.
$$
Therefore $D_{a,n}=\det M=(1-a^2)^n$.
</s>