Find the determinant of a Vandermonde-like matrix I reduced a certain determinant to the following which has the form:
\begin{vmatrix}
1&1&\cdots & 1 \\
-a_1&a_2&\cdots &a_n\\
-a_1^{2}&a_2^{2}&\cdots&a_n^{2}\\
\vdots&\vdots&\ddots& \vdots\\
-a_1^{n-1}&a_2^{n-1}&\cdots&a_n^{n-1}\\
\end{vmatrix}
To clarify a bit, it is exactly the Vandermonde determinant except the first column is negative, while the $(1,1)$-entry is still $1$. 
I think the key is to apply Vandermonde determinant yet I can't proceed. There may be a quick answer to this, however. Any hints?

By linearity in or Laplace expansion along the first column, the determinant is equal to
$$
\begin{vmatrix}
-1&1&\cdots & 1 \\
-a_1&a_2&\cdots &a_n\\
-a_1^{2}&a_2^{2}&\cdots&a_n^{2}\\
\vdots&\vdots&\ddots& \vdots\\
-a_1^{n-1}&a_2^{n-1}&\cdots&a_n^{n-1}\\
\end{vmatrix}
+
2\begin{vmatrix}
a_2&\cdots &a_n\\
a_2^{2}&\cdots&a_n^{2}\\
\vdots&\ddots& \vdots\\
a_2^{n-1}&\cdots&a_n^{n-1}\\
\end{vmatrix}.
$$
Now each of the two summands is a multiple of a Vandermonde determinant.
