Calculate the determinant of
$$ \left[ \begin{array}{cccc} 1 & 0 & 0 & 0 & \cdots & 1\\ 1 & a_1 & 0 & 0 & \cdots & 0 \\ 1 & 1 & a_2 & 0 & \cdots & 0 \\ 1 & 0 & 1 & a_3 & \cdots & 0 \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\ 1 & 0 & 0 & \cdots& 1 & a_{n-1} \\ \end{array} \right]$$
I tried to develop at the first line, but got stuck. Any helps or hints appreciated.
Expanding from the first one will force a product along the leading diagonal; the one at the end of the first row will leave you with another determinant \begin{eqnarray*} \left[ \begin{array}{cccc} 1 & 0 & 0 & 0 & \cdots & 1\\ 1 & a_1 & 0 & 0 & \cdots & 0 \\ 1 & 1 & a_2 & 0 & \cdots & 0 \\ 1 & 0 & 1 & a_3 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \ddots & \vdots \\ 1 & 0 & 0 & \cdots& 1 & a_{n-1} \\ \end{array} \right]=a_1 \cdots a_{n-1}+(-1)^{n+1}\left[ \begin{array}{cccc} 1 & a_1 & 0 & 0 & \cdots & 0 \\ 1 & 1 & a_2 & 0 & \cdots & 0 \\ 1 & 0 & 1 & a_3 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \ddots & \vdots \\ 1 & 0 & 0 & \cdots& 1 & a_{n-2} \\ 1 & 0 & 0 & \cdots& 0 & 1 \\ \end{array} \right] \end{eqnarray*} Now take the last row of this determinant & make it the first row & shift all the other rows down by one; this will introduce another $(-1)^{n+1}$ but will be essentially the same determinant that you started with but of size one smaller \begin{eqnarray*} -(-1)^{n+1}(-1)^{n+1}\left[ \begin{array}{cccc} 1 & 0 & 0 & 0 & \cdots & 1\\ 1 & a_1 & 0 & 0 & \cdots & 0 \\ 1 & 1 & a_2 & 0 & \cdots & 0 \\ 1 & 0 & 1 & a_3 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \ddots & \vdots \\ 1 & 0 & 0 & \cdots& 1 & a_{n-2} \\ \end{array} \right] \end{eqnarray*} So now we can use the above arguement recursively & we have \begin{eqnarray*} D=a_1 \cdots a_{n-1}-a_1 \cdots a_{n-2} +\cdots (-1)^{n-2}a_1+ (-1)^{n-1}. \end{eqnarray*} Tidy this expression up a little bit & we have the answer (exactly as Arden states in his comment) \begin{eqnarray*} D=\sum_{k=0}^{n-1}(-1)^{n-1-k}\prod_{i=1}^ka_i. \end{eqnarray*}