Calculate the determinant Calculate the determinant

$$\begin{align*}D[n]=\begin{array}{cccccc} b & b & b & \dots & b & a \\ b & b & b & \dots & a & b \\ \vdots & \vdots & \vdots &  & \vdots & \vdots \\ b & b & a & b & b & b \\ b & a & b & b & b & b \\ a & b & b & b & b & b \\\end{array}\end{align*}$$
$$\begin{align*}\left(\begin{array}{ccccc} b & b & b & b & a \\ b & b & b & a & b \\ b & b & a & b & b \\ b & a & b & b & b \\ a & b & b & b & b \\\end{array}\right)\end{align*}$$
I calculated the value of $n=2$, and $n=3$, $(a-b)^2(a+2 b)$
I guess the result is $(a-b)^{n-1}(a+(n-1)b)(-1)^{n-1}$
So, I need a prove by induction? how to do that?
I'm not sure whether this is easier to do than think out one new method.

You may swap the rows to bring the $a$s to the main diagonal (what effect does that have on the sign?).
Then replace $a$ with $a-X$ to obtain the characteristic polynomial $\chi_M(X)$ of your matrix while we are looking for $\chi_M(0)$. Its roots are the eigenvalues with corresponding multiplicity. Note that $(1\;1\;\ldots\;1)$ is an obvious eigenvector of eigenvalue $a+(n-1)b$ and the $(n-1)$-dimensional space orthogonal to it consists of eigenvectors of eigenvalue $a-b$.
Hence your conjecture follows.
