In the following let $J$ denote an ordered $k$-tuple $\{j_1, \ldots, j_k\}$, $1\leq j_1 <\cdots < j_k \leq n$, and $s$, $1\leq s\leq n$ such that $s\neq j_1,\ldots, j_k $. If $\phi_1,\ldots, \phi_k$ are $k$ smooth functions of $x_1, \ldots, x_n$ and $\dfrac{\partial(\phi_1, \ldots, \phi_k)}{\partial(x_{j_1}, \ldots, x_{j_k})}$ denote the determinant of the Jacobian of the map $\phi =(\phi_1, \ldots, \phi_k)$, then $$\sum_{s,J} \dfrac{\partial}{\partial x_s}\dfrac{\partial(\phi_1, \ldots, \phi_k)}{\partial(x_{j_1}, \ldots, x_{j_k})} dx_s \wedge dx_{j_1} \wedge \cdots \wedge dx_{j_k} =0.$$ My idea is to form up complementary pairs so that we can have an internal cancellation. But how do we get the corresponding pairs of indices who will cancel each other. Please help me.
To make sure we can form up pairs we have to have the total number of terms in the sum should be even. I have worked out the total number of terms in the sum is $n_{C_k} \times (n-k) \times k$, and I believe that this is even.
If anyone have a better idea how to prove this sum is zero I will be grateful. Thankyou in advance.