SFAICT this route hasn't been mentioned yet, so:
Consider the decomposition
$$\small\begin{pmatrix}a&b&\cdots&b\\b&a&\cdots&b\\\vdots&&\ddots&\vdots\\b&\cdots&b&a\end{pmatrix}=\begin{pmatrix}a-b&&&\\&a-b&&\\&&\ddots&\\&&&a-b\end{pmatrix}+\begin{pmatrix}\sqrt b\\\sqrt b\\\vdots\\\sqrt b\end{pmatrix}\cdot\begin{pmatrix}\sqrt b&\sqrt b&\cdots&\sqrt b\end{pmatrix}$$
Having this decomposition allows us to use the Sherman-Morrison-Woodbury formula for determinants:
$$\det(\mathbf A+\mathbf u\mathbf v^\top)=(1+\mathbf v^\top\mathbf A^{-1}\mathbf u)\det\mathbf A$$
where $\mathbf u$ and $\mathbf v$ are column vectors. The corresponding components are simple, and thus the formula is easily applied (letting $\mathbf e$ denote the column vector whose components are all $1$'s):
$$\begin{align*}
\begin{vmatrix}a&b&\cdots&b\\b&a&\cdots&b\\\vdots&&\ddots&\vdots\\b&\cdots&b&a\end{vmatrix}&=\left(1+(\sqrt{b}\mathbf e)^\top\left(\frac{\sqrt{b}}{a-b}\mathbf e\right)\right)(a-b)^n\\
&=\left(1+\frac{nb}{a-b}\right)(a-b)^n=(a+(n-1)b)(a-b)^{n-1}
\end{align*}$$
where we used the fact that $\mathbf e^\top\mathbf e=n$.