Subtracting the last row of blocks from the first $k-1$ rows of blocks, we obtain

$$\begin{bmatrix}A-B & O & O & \dots & O & B-A\\ O & A-B & O & \dots & O & B-A\\ O & O & A-B & \dots & O & B-A\\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ O & O & O & \dots & A-B & B-A\\ B & B & B & \dots & B & A\end{bmatrix}$$

which is the first step in Remling's answer. Adding the first $k-1$ columns of blocks to the last column of blocks, we obtain a block lower triangular matrix

$$\begin{bmatrix}A-B & O & O & \dots & O & O\\ O & A-B & O & \dots & O & O\\ O & O & A-B & \dots & O & O\\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ O & O & O & \dots & A-B & O\\ B & B & B & \dots & B & M_k (A,B)\end{bmatrix}$$

where $M_k (A,B) = A+(k-1)B$. Adding the first $k-1$ rows of blocks to the last row of blocks multiplied by $k$, we obtain

$$\begin{bmatrix}A-B & O & O & \dots & O & O\\ O & A-B & O & \dots & O & O\\ O & O & A-B & \dots & O & O\\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ O & O & O & \dots & A-B & O\\ M_k (A,B) & M_k (A,B) & M_k (A,B) & \dots & M_k (A,B) & k \,M_k (A,B)\end{bmatrix}$$

Unlike the previous two row / column operations, which preserved the determinant, this one multiplies the determinant by $k^n$. We can divide the determinant by $k^n$ by multiplying the last column of blocks by $k^{-1}$, which produces

$$\begin{bmatrix}A-B & O & O & \dots & O & O\\ O & A-B & O & \dots & O & O\\ O & O & A-B & \dots & O & O\\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ O & O & O & \dots & A-B & O\\ M_k (A,B) & M_k (A,B) & M_k (A,B) & \dots & M_k (A,B) & M_k (A,B)\end{bmatrix}$$

Subtracting the last column of blocks from the first $k-1$ columns of blocks, we obtain the following block diagonal matrix

$$\begin{bmatrix} A-B & O & O & \dots & O & O\\ O & A-B & O & \dots & O & O\\ O & O & A-B & \dots & O & O\\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ O & O & O & \dots & A-B & O\\ O & O & O & \dots & O & M_k (A,B)\end{bmatrix}$$

whose determinant is

$$(\det(A-B))^{k-1} \det(M_k (A,B)) = (\det(A-B))^{k-1} \det (A + (k-1) B)$$

Thus,

$$\det(C) = (\det(A-B))^{k-1} \det (A + (k-1) B)$$

Subtracting the last row of blocks from the first $k-1$ rows of blocks, we obtain

$$\begin{bmatrix}A-B & O & O & \dots & O & B-A\\ O & A-B & O & \dots & O & B-A\\ O & O & A-B & \dots & O & B-A\\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ O & O & O & \dots & A-B & B-A\\ B & B & B & \dots & B & A\end{bmatrix}$$

which is the first step in Remling's answer. Adding the first $k-1$ columns of blocks to the last column of blocks, we obtain a block lower triangular matrix

$$\begin{bmatrix}A-B & O & O & \dots & O & O\\ O & A-B & O & \dots & O & O\\ O & O & A-B & \dots & O & O\\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ O & O & O & \dots & A-B & O\\ B & B & B & \dots & B & M_k (A,B)\end{bmatrix}$$

where $M_k (A,B) = A+(k-1)B$. The determinant of this block lower triangular matrix is

$$(\det(A-B))^{k-1} \det(M_k (A,B)) = (\det(A-B))^{k-1} \det (A + (k-1) B)$$

Thus,

$$\det(C) = (\det(A-B))^{k-1} \det (A + (k-1) B)$$

Subtracting the last row of blocks from the first $k-1$ rows of blocks, we obtain

$$\begin{bmatrix}A-B & O & O & \dots & O & B-A\\ O & A-B & O & \dots & O & B-A\\ O & O & A-B & \dots & O & B-A\\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ O & O & O & \dots & A-B & B-A\\ B & B & B & \dots & B & A\end{bmatrix}$$

which is the first step in Remling's answer. Adding the first $k-1$ columns of blocks to the last column of blocks, we obtain a block lower triangular matrix

$$\begin{bmatrix}A-B & O & O & \dots & O & O\\ O & A-B & O & \dots & O & O\\ O & O & A-B & \dots & O & O\\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ O & O & O & \dots & A-B & O\\ B & B & B & \dots & B & M_k (A,B)\end{bmatrix}$$

where $M_k (A,B) = A+(k-1)B$. Adding the first $k-1$ rows of blocks to the last row of blocks multiplied by $k$, we obtain

$$\begin{bmatrix}A-B & O & O & \dots & O & O\\ O & A-B & O & \dots & O & O\\ O & O & A-B & \dots & O & O\\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ O & O & O & \dots & A-B & O\\ M_k (A,B) & M_k (A,B) & M_k (A,B) & \dots & M_k (A,B) & k \,M_k (A,B)\end{bmatrix}$$

Unlike the previous two row / column operations, this one multiplies the determinant by $k$. We can divide the determinant by $k$ by multiplying the last column of blocks by $k^{-1}$, which produces

$$\begin{bmatrix}A-B & O & O & \dots & O & O\\ O & A-B & O & \dots & O & O\\ O & O & A-B & \dots & O & O\\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ O & O & O & \dots & A-B & O\\ M_k (A,B) & M_k (A,B) & M_k (A,B) & \dots & M_k (A,B) & M_k (A,B)\end{bmatrix}$$

Subtracting the last column of blocks from the first $k-1$ columns of blocks, we obtain the following block diagonal matrix

$$\begin{bmatrix} A-B & O & O & \dots & O & O\\ O & A-B & O & \dots & O & O\\ O & O & A-B & \dots & O & O\\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ O & O & O & \dots & A-B & O\\ O & O & O & \dots & O & M_k (A,B)\end{bmatrix}$$

whose determinant is

$$(\det(A-B))^{k-1} \det(M_k (A,B)) = (\det(A-B))^{k-1} \det (A + (k-1) B)$$

Thus,

$$\det(C) = (\det(A-B))^{k-1} \det (A + (k-1) B)$$

Subtracting the last row of blocks from the first $k-1$ rows of blocks, we obtain

$$\begin{bmatrix}A-B & O & O & \dots & O & B-A\\ O & A-B & O & \dots & O & B-A\\ O & O & A-B & \dots & O & B-A\\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ O & O & O & \dots & A-B & B-A\\ B & B & B & \dots & B & A\end{bmatrix}$$

which is the first step in Remling's answer. Adding the first $k-1$ columns of blocks to the last column of blocks, we obtain a block lower triangular matrix

$$\begin{bmatrix}A-B & O & O & \dots & O & O\\ O & A-B & O & \dots & O & O\\ O & O & A-B & \dots & O & O\\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ O & O & O & \dots & A-B & O\\ B & B & B & \dots & B & M_k (A,B)\end{bmatrix}$$

where $M_k (A,B) = A+(k-1)B$. Adding the first $k-1$ rows of blocks to the last row of blocks multiplied by $k$, we obtain

$$\begin{bmatrix}A-B & O & O & \dots & O & O\\ O & A-B & O & \dots & O & O\\ O & O & A-B & \dots & O & O\\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ O & O & O & \dots & A-B & O\\ M_k (A,B) & M_k (A,B) & M_k (A,B) & \dots & M_k (A,B) & k \,M_k (A,B)\end{bmatrix}$$

Unlike the previous two row / column operations, which preserved the determinant, this one multiplies the determinant by $k^n$. We can divide the determinant by $k^n$ by multiplying the last column of blocks by $k^{-1}$, which produces

$$\begin{bmatrix}A-B & O & O & \dots & O & O\\ O & A-B & O & \dots & O & O\\ O & O & A-B & \dots & O & O\\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ O & O & O & \dots & A-B & O\\ M_k (A,B) & M_k (A,B) & M_k (A,B) & \dots & M_k (A,B) & M_k (A,B)\end{bmatrix}$$

Subtracting the last column of blocks from the first $k-1$ columns of blocks, we obtain the following block diagonal matrix

$$\begin{bmatrix} A-B & O & O & \dots & O & O\\ O & A-B & O & \dots & O & O\\ O & O & A-B & \dots & O & O\\ \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ O & O & O & \dots & A-B & O\\ O & O & O & \dots & O & M_k (A,B)\end{bmatrix}$$

whose determinant is

$$(\det(A-B))^{k-1} \det(M_k (A,B)) = (\det(A-B))^{k-1} \det (A + (k-1) B)$$

Thus,

$$\det(C) = (\det(A-B))^{k-1} \det (A + (k-1) B)$$