My homework is to calculate the determinant: $$ \det(A) = \begin{vmatrix} a_1 & a_1 & a_1 & \cdots & a_1 \\\ a_1 & a_2 & a_2 & \cdots & a_2 \\\ \vdots & \vdots & \vdots & \ddots & \vdots \\\ a_1 & a_2 & a_3 & \cdots & a_n \end{vmatrix} = V(a_1, a_2, \dots, a_n)$$ and following is my proof.
Expand $A$ using the first line:
$$ \det(A) = (-1)^{1+1} a_1 \begin{vmatrix} a_2 & a_2 & \cdots & a_2 \\\ a_2 & a_3 & \cdots & a_3 \\\ \vdots & \vdots & \ddots & \vdots \\\ a_2 & a_3 & \cdots & a_n \end{vmatrix} + (-1)^{1+2} a_1 \begin{vmatrix} a_1 & a_2 & \cdots & a_2 \\\ a_1 & a_3 & \cdots & a_3 \\\ \vdots & \vdots & \ddots & \vdots \\\ a_1 & a_3 & \cdots & a_n \end{vmatrix} + (-1)^{1+3} a_1 \begin{vmatrix} a_1 & a_2 & \cdots & a_2 \\\ a_1 & a_2 & \cdots & a_3 \\\ \vdots & \vdots & \ddots & \vdots \\\ a_1 & a_2 & \cdots & a_n \end{vmatrix} + \dots + (-1)^{1+n} a_1 \begin{vmatrix} a_1 & a_2 & \cdots & a_2 \\\ a_1 & a_2 & \cdots & a_3 \\\ \vdots & \vdots & \ddots & \vdots \\\ a_1 & a_2 & \cdots & a_{n-1} \end{vmatrix} $$
Or $$ \det(A) = a_1 \det(A_{1,1}) - a_1 \det(A_{1,2}) + \sum\limits_{i=3}^{n} a_1 (-1)^{1+i} \det(A_{1,i}) $$
If $i \geq 3$, the first two columns of $A_{1,i}$ are proportional, then $\det(A_{1,i})=0$. Moreover $$ \det(A_{1,1}) = V(a_2,a_3,\dots,a_n) $$ and $$ \det(A_{1,2}) = \frac{a_1}{a_2} V(a_2,a_3,\dots,a_n) $$ So $$ \begin{align} V(a_1, a_2, \dots, a_n) & = a_1 V(a_2,a_3,\dots,a_n) - a_1 \frac{a_1}{a_2} V(a_2,a_3,\dots,a_n) \\\ & = a_1 \frac{a_2 - a_1}{a_2} V(a_2,a_3,\dots,a_n) \\\ & = a_1 \frac{a_2 - a_1}{a_2} a_2 \frac{a_3 - a_2}{a_3} V(a_3,\dots,a_n) \end{align} $$ Since $V(a_n) = a_n$, $$ V(a_1, a_2, \dots, a_n) = a_1 \frac{a_2 - a_1}{a_2} a_2 \frac{a_3 - a_2}{a_3} \dots a_{n-1} \frac{a_n - a_{n-1}}{a_n} a_{n} = a_{1} \prod_{i=2}^{n} (a_i - a_{i-1}) $$
For the feedback, I've got a simple remark "wrong" by the grader, but I still cannot find where the error is.
Thanks for any help.