I have the following $N \times N$ matrix.
\begin{vmatrix}
0 & 1 & 1 & \ldots & 1 \\
1 & a_1 & 0 & \ldots & 0 \\
1 & 0 & a_2 & \ldots & 0 \\
\vdots & \vdots& &\ddots& \vdots\\
1 & 0 & 0 & \ldots & a_n \\
\end{vmatrix}
There seems to be a pattern going on for the determinant of the $5 \times 5$ version of this matrix, but I'm not sure how I would find the determinant for the $N \times N$ one.
Answer
Transform the matrix by the (determinant invariant) operation of adding $-a_i$ times the $(i+1)$th row on the first row. This gets us
\begin{vmatrix}
-\sum_i \frac{1}{a_i} & 0 & 0 & \ldots & 0 \\
1 & a_1 & 0 & \ldots & 0 \\
1 & 0 & a_2 & \ldots & 0 \\
\vdots & \vdots& &\ddots& \vdots\\
1 & 0 & 0 & \ldots & a_n \\
\end{vmatrix}
Then you have a lower triangle matrix whose determinant is just the product of the diagonal elements.
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