Proof of Active Coordinate Coefficient Formula for the Exterior Derivative

Write

By Exterior Derivative of a C^1 Differential Form on a Euclidean Open Set, one has

We determine the coefficient of $dx_{i_1}\wedge\cdots\wedge dx_{i_k}$. A summand can contribute to this coefficient only if the set of indices in

is exactly $\{i_1,\dots,i_k\}$. Since the tuple $(j_1,\dots,j_{k-1})$ is increasing, this happens precisely when $(j_1,\dots,j_{k-1})=I^{\hat r}$ and $m=i_r$ for some $r\in\{1,\dots,k\}$.

Fix such an index $r$. Then

Let $\sigma_r\in S_k$ be the permutation that moves $i_r$ from the first position to the $r$th position and shifts $i_1,\dots,i_{r-1}$ one place to the left. This permutation has sign $(-1)^{r-1}$. Hence by Permutation Rule for Wedge Products of Coordinate 1-Forms,

Therefore the total coefficient of $dx_{i_1}\wedge\cdots\wedge dx_{i_k}$ in $d\omega$ is exactly