Proof for the determinant of a Cauchy matrix

Question

I want to proof the formula for the determinant of a Cauchy Matrix without recurring to matrix manipulation, but by directly applying the definition of the determinant. That is, given two sequences of numbers of length n, $x_1,...,x_n$ and $y_1,...,y_n$, where $x_i \neq -y_i$, I want to show that the determinant of the matrix A, whose entries in the ith row and jth column are given by \begin{align} \frac{1}{x_i+y_j} \end{align}, is given by: \begin{align} \det(A) = \frac{\prod_{i<j} (x_i-x_j)(y_i-y_j)}{\prod_{i,j} x_i+x_j} \end{align} Now my proof looks like this: \begin{align} \det(A) &= \sum_{\delta \in perm(N)} sign(\delta)\prod_j \frac{1}{x_{\delta j}+y_j}\\ & = \frac{\sum_{\delta} -sign(\delta)\prod_j x_{\delta j}+y_j}{\prod_\delta \prod_j x_{\delta j} +y_j} \\&= \quad ...\\ &= \frac{\left (\frac {\prod_\delta \prod_j x_{\delta j} +y_j}{\prod_{i,j} x_i+x_j} \right ) \prod_{i<j} (x_i-x_j)(y_i-y_j) }{\prod_\delta \prod_j x_{\delta j} +y_j}\\&=\frac{\prod_{i<j} (x_i-x_j)(y_i-y_j)}{\prod_{i,j} x_i+x_j} \end{align} Where $N$ is the set of the first $n$ natural numbers. What I have yet to understand, is the jump from equation 2 to 4. For the case of $n = 2$ it obviously holds. For the case of $n>2$ my intuition says that for every permutation there is a permutation with opposite sign such that $\prod_j x_{\delta j}+y_j$ has $n-2$ common factors and that these can be factored out such that they cancel with the $x_i+x_j$ appearing more than once in $\prod_\delta \prod_j x_{\delta j} +y_j$. Can you help me to formalize that intuition or, if its wrong, lead me towards the right path?