Determinant of $A_{2n}$ Determine the determinant of $$A_{2n}:=\begin{pmatrix}
a &0  &\cdots   & \cdots &0  &b \\ 
 0& a &  &  &  b&0 \\ 
 \vdots &\vdots  &\ddots   &\ddots  &\vdots  &\vdots \\ 
 \vdots&\vdots  &  \ddots & \ddots  &\vdots  &\vdots  \\ 
 0& b &  &  &  a&0 \\ 
 b&0  &\cdots   &\cdots   &0  &a 
\end{pmatrix}\in \mathbb{R}^{2n \times 2n}$$
My initial thought is if we assume that it is $2 \times 2$, then we will get $a^2-b^2$. If it is a $4 \times 4$ one then we would get $(a^2-b^2)^2$. So would the determinant for a $2n \times 2n$ is $(a^2-b^2)^n$? But how to prove it mathematically?

$\vert A_{2n}\vert=\begin{vmatrix} B_{n} & C_{n} \\ C_{n} & B_{n} \end{vmatrix}$, Where $B_{n}=aI_{n}$ and $C_{n}=bI_{n}$. Because $B_{n}$ and $C_{n}$ non singular, then
$\vert A_{2n}\vert=\vert B_{n}^2-C_{n}^2\vert=\vert a^2I_{n}-b^2I_{n}\vert=\vert (a^2-b^2)I_{n}\vert=(a^2-b^2)^n.1=(a^2-b^2)^n.$
$\blacksquare$
