How to find the determinant of this $n \times n$ matrix in a clever way? Is there any clever and short way to find out the determinant of the following matrix?
\begin{bmatrix}
        b_1 & b_2 & b_3 & \cdots & b_{n-1} & 0 \\
        a_1 & 0 & 0 & \cdots & 0 & b_1 \\
        0 & a_2 & 0 & \cdots & 0 & b_2\\
        \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
        0 & 0 & 0 & \cdots & a_{n-1} & b_{n-1} \\
        \end{bmatrix}
Any help will be appreciated.

$b_1 \det\begin{pmatrix}
        0 & 0 & \cdots & 0 & b_1 \\
        a_2 & 0 & \cdots & 0 & b_2\\
        \vdots & \vdots & \vdots & \ddots & \vdots \\
        0 & 0 & \cdots & a_{n-1} & b_{n-1} \\
        \end{pmatrix} 
- a_1 \det\begin{pmatrix}
        b_2 & b_3 & \cdots & b_{n-1} & 0 \\
        a_2 & 0 & \cdots & 0 & b_2\\
        \vdots & \vdots & \vdots & \ddots & \vdots \\
        0 & 0 & \cdots & a_{n-1} & b_{n-1} \\
        \end{pmatrix}$
$(-1)^nb_1\cdot a_2\cdot a_3\cdots a_{n-1}\cdot b_1-\\
 a_1\cdot b_2 \det\begin{pmatrix}
        0 & 0 & \cdots & 0 & b_2 \\
        a_3 & 0 & \cdots & 0 & b_3\\
        \vdots & \vdots & \vdots & \ddots & \vdots \\
        0 & 0 & \cdots & a_{n-1} & b_{n-1} \\
        \end{pmatrix}+\\
a_1\cdot a_2 \det\begin{pmatrix}
        b_3 & b_4 & \cdots & b_{n-1} & 0 \\
        a_3 & 0 & \cdots & 0 & b_3\\
        \vdots & \vdots & \vdots & \ddots & \vdots \\
        0 & 0 & \cdots & a_{n-1} & b_{n-1} \\
        \end{pmatrix}$ 
I think we have a pattern.
$b_1^2\cdot a_2\cdots a_{n-1} + a_1\cdot b_2^2\cdot a_3\cdots a_{n-1}+a_1\cdot a_2\cdot b_3^2\cdot a_4\cdots a_{n-1}+\cdots$
$(-1)^n\sum_\limits{j=1}^{n-1} b_j^2 \frac{\prod_\limits{k=1}^{n-1} a_k}{a_j}$ 
