Seeking advice on how to tackle the below determinant
$$ \begin{vmatrix} a_1b_1 & a_1b_2 & a_1b_3 & \ldots & a_1b_n \\ a_1b_2 & a_2b_2 & a_2b_3 & \ldots & a_2b_n \\ a_1b_3 & a_2b_3 & a_3b_3 & \ldots & a_3b_n \\ \ldots & \ldots & \ldots & \ddots & \vdots\\ a_1b_n & a_2b_n & a_3b_n & \ldots & a_nb_n\end{vmatrix} $$
I have tried countless techniques, including cofactor expansion as well as row/column swapping. I also tried to exploit the symmetricness of the matrix. I have looked elsewhere for guidance however this is apparently quite rare. Any help or advice would be appreciated.
The answer is $\color{blue}{a_1 b_n\prod_{k=1}^{n-1}(a_{k+1}b_k-a_k b_{k+1})}$. Indeed, call it $D_n$ and assume $n>1$. If we multiply the $n$-th (i.e. last) column of the matrix by $b_{n-1}$, and subtract the $(n-1)$-th column multiplied by $b_n$, we get a column with the last entry equal to $b_n(a_n b_{n-1}-a_{n-1}b_n)$, and all the other entries equal to zero. Hence $b_{n-1}D_n=b_n(a_n b_{n-1}-a_{n-1}b_n)D_{n-1}$. Using the fact that $D_n$ is a polynomial in $a_k$ and $b_k$ (to handle cases of $b_k$ being zero), we finish by induction.