Valid induction? The problem was presented like this:
$A_{2n} = 
\begin{bmatrix}
a & 0 & \cdots & 0 & b\\ 
0 & a & \cdots & b & 0\\ 
 &  & \cdots &  & \\ 
0 & b & \cdots & a & 0\\ 
b & 0 & \cdots & 0 & a
\end{bmatrix} 
n \in N, A_{0} = 1
$
Show that $detA_{2n} = (a^{2} - b^{2}) \cdot detA_{2n-2}$
I tried to use induction for this case:
Base case:
$n=1$: $detA_{2n} = \begin{vmatrix}a & b\\ b & a\end{vmatrix} = a^{2}-b^{2} = a^{2}-b^{2} \cdot detA_{2n-2} = a^{2}-b^{2} \cdot detA_{0} = a^{2}-b^{2} \cdot 1 = a^{2}-b^{2} $
Hypothesis:
$ detA_{2n} = (a^{2} - b^{2}) \cdot detA_{2n-2} \Rightarrow detA_{2(n+1)} = (a^{2} - b^{2}) \cdot detA_{2(n+1)-2} $ 
$detA_{2(n+1)} = (a^{2} - b^{2}) \cdot detA_{2(n+1)-2} \Leftrightarrow detA_{2n+2} = (a^{2} - b^{2}) \cdot detA_{2n+2-2} \Leftrightarrow detA_{2n+2} = (a^{2} - b^{2}) \cdot detA_{2n} \Leftrightarrow detA_{2n} = (a^{2} - b^{2}) \cdot detA_{2n-2}$
This feels circular, is this the right way to do it?

$$A_{2(n+1)} = 
\begin{bmatrix}
a & 0 & \cdots & 0 & b\\ 
0 & a & \cdots & b & 0\\ 
 &  & \cdots &  & \\ 
0 & b & \cdots & a & 0\\ 
b & 0 & \cdots & 0 & a
\end{bmatrix}.$$
We will proceed using the appropriate cofactor matrices. First, expand along the first row, note that we have to negate the $(1,2n+2)$-th entry b since $2n+3$ is odd: 
$$\det(A_{2(n+1)})=a \det\begin{bmatrix}
 a & \cdots & b & 0\\ 
 &  & \cdots &  & \\ 
 b & \cdots & a & 0\\ 
 0 & \cdots & 0 & a
\end{bmatrix}-b\det\begin{bmatrix}
0 & a & \cdots & b \\ 
 &  & \cdots &  & \\ 
0 & b & \cdots & a \\ 
b & 0 & \cdots & 0 
\end{bmatrix}.$$
Here the cofactor matrices are of size (2n+1) by (2n+1). Now we expand the first resulting cofactor matrix along the last row and the second cofactor matrix along the first column. By the same logic as before, we don't have to negate anything in this case:
$$\det(A_{2(n+1)})=a^2\det(A_{2n})-b^2\det(A_{2n})=(a^2-b^2)\det(A_{2n}).$$
Hence we have finished the second part of the induction. Combine it with the base case check to complete the proof.
