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?

We expand the determinant along the first row:
$$
\det A_{2n}=aM_{11}+(-1)^{2n+1}bM_{1,2n}
$$
Now we expand $M_{11}$ and $M_{1,2n}$ along the last row:
$$
\det A_{2n}=a^2\det A_{2n-2}-b^2\det A_{2n-2}=(a^2-b^2)\det A_{2n-2}.
$$
Now induction works:
$$
\det A_{2n}=(a^2-b^2)(a^2-b^2)^{n-1}=(a^2-b^2)^n.
$$
