Note that if we take $(\bmod 2)$, we get that $\prod_{1 \leq i<j \leq n}\left(x_{i}+x_{j}\right) \equiv \prod_{1 \leq i<j \leq n}\left(x_{j}-x_{i}\right)=\operatorname{det}(M)$ where $M$ is the matrix with $M_{ij}=x_{i}^{j-1}$. This is called a Vandermonde determinant. Expanding this determinant using the formula $\operatorname{det}(M)=\sum_{\sigma} \prod_{i=1}^{n} x_{\sigma(i)}^{i-1}$ where the sum if over all $n$! permutations $\sigma$, gives the result.