Solving Vandermonde equation system Given the following:

$$\begin{pmatrix} 1 & 1 & 1 & ... & 1 \\ a_1 & a_2 & a_3 & ... & a_n \\ a_1^2 & a_2^2 & a_3^2 & ... & a_n^2 \\ \vdots & \vdots & \vdots & \vdots & \vdots \\  a_1^{n-1} & a_2^{n-1} & a_3^{n-1} & ... & a_n^{n-1}\end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ \vdots \\ x_n\end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \\ \vdots \\ 1 \end{pmatrix} $$

How can this be solved to give $x_k$? By Cramer's rule, this should be $x_k=\frac{\det(A_i)}{\det(A)}$
How can this result be simplified? Since this is a Vandermonde matrix, $\det(A)=\prod_{1\leq i<j\leq n}(a_j-a_i)$. However, can the numerator be written any simpler? In a paper where I have seen this equation system used, the result is given as: $x_k=\prod_{1\leq j \leq n,\;j \neq i} (a_i - a_j)^{-1}$ but it's not clear how the simplification has been achieved?
 A: Using Cramer's rule, we know
$$x_k = \frac{\det(A_k)}{\det(A)}$$
Now, as we know the Vandermonde determinant is equal to
$$\det(A)=\prod_{1\leq i<j\leq n}(a_j-a_i)$$
To calculate $\det(A_k)$, we replace the kth column with $b_i$s which in this case are all $0$s and last entry $1$. If we calculate the determinant using the kth column, we need only consider the determinant using the last entry as all the other ones will multiply with $0$ and become $0$. This last entry sub determinant also turns out to be a vandermonde determinant of one lower degree. So calculating the determinant using that 
$$\det(A_k) = (-1)^{n+k}\prod_{1\leq i<j\leq n}(a_j-a_i)$$
where both $i,j\neq k$
When you divide them, all the terms that include $a_k$ are all that are left.
$$x_k = \left[(-1)^{n+k}\left(\prod_{k+1\leq j\leq n}(a_j-a_k) \prod_{1\leq i\leq k-1}(a_k-a_i)\right)\right]^{-1}$$
The terms which have greater index than $a_k$ come under one umbrella with the greater index term leading and the ones with lower come under a second umbrella with $a_k$ leading.
Rearrange all the terms in first umbrella so that $a_k$ term is leading and take all the $-1$s that come out and multiply it with existing to get your answer .
$$x_k = \left[(-1)^{n+k}\left((-1)^{n-(k+1)}\prod_{k+1\leq j\leq n}(a_k-a_j) \prod_{1\leq i\leq k-1}(a_k-a_i)\right)\right]^{-1}$$
$$x_k = \left[(-1)^{n+k+n-k-1}\left(\prod_{k+1\leq j\leq n}(a_k-a_j) \prod_{1\leq i\leq k-1}(a_k-a_i)\right)\right]^{-1}$$
$$x_k = \left[(-1)^{2n-1}\left(\prod_{k+1\leq j\leq n}(a_k-a_j) \prod_{1\leq i\leq k-1}(a_k-a_i)\right)\right]^{-1}$$
$$x_k = \left[\prod_{1\leq i\leq n, i\neq k}(a_i-a_k)\right]^{-1}$$
