Let $\Omega$ be a open subset of $\mathbb{R}^{N}$ and $\omega: X(\Omega) \times \cdots \times X(\Omega) \to C^\infty(\Omega)$ be a $k$-form, that is, $\omega$ is $C^\infty(\Omega)$-alternating multilinear map, where $X(\Omega)$ denotes the space of vector fields in $\Omega$, that is, $X(\Omega)=\{L:C^\infty(\Omega) \to C^\infty(\Omega):L \hbox{ is $\mathbb{R}$-linear and } L(fg)=fL(g)+gL(f)\}$.
It is possible to show that $L=\sum_{j=1}^N a_j(x) \frac{\partial}{\partial x_j}$.
By an ordered multi-index of length $k$ we mean a finite sequence of the form $J=\{j_1, \dots, j_k\}$ with $1 \leq j_1<\cdots<j_k \leq N$ and we write $|J|=k$.
Define $$\omega_J=\omega\left(\frac{\partial}{\partial x_{j_1}},\dots, \frac{\partial}{\partial x_{j_k}} \right)\in C^\infty(\Omega), ~ J=\{j_1, \dots, j_k\}$$ and $dx_J$ be the $k-$form given by $$dx_J\left(\frac{\partial}{\partial x_{i_1}},\dots, \frac{\partial}{\partial x_{i_k}} \right)=\delta_{IJ}, I=\{i_1, \dots, i_k\}.$$
My question: How to prove that $\omega=\sum_{|J|=k}^{'}\omega_J dx_J$, where $\sum_{|J|=k}^{'}$ indicates that the sum is given only on the ordered multi-indexes of length $k$?
My attempt: I tried to use induction in $k$. For $k=1$, $L \in X(\Omega)$ and $b_j=\omega\left(\frac{\partial}{\partial x_j}\right)$ we have:
\begin{eqnarray} \omega(L)&=& \omega\left( \sum_{j=1}^{N} a_j \frac{\partial}{\partial x_j} \right)\\ &=& \sum_{j=1}^{N}a_jb_j\\ &=& \sum_{j=1}^{N} a_j \left( \sum_{k=1}^{N} b_kdx_k \left(\frac{\partial}{\partial x_j}\right)\right)\\ &=& \sum_{j=1}^{N} \sum_{k=1}^{N} a_j b_k dx_k\left(\frac{\partial}{\partial x_j} \right)\\ &=& \left( \sum_{k=1}^{N} b_k dx_k \right) \left( \sum_{j=1}^{N} a_j \frac{\partial}{\partial x_j} \right)\\ &=& \left(\sum_{k=1}^{N} b_k dx_k \right) (L)\\ &=&\left(\sum_{|J|=1}'\omega_Jdx_J \right)(L), \end{eqnarray} since $dx_j(L)=L(x_j)$ and thus, $dx_i\left(\frac{\partial}{\partial x_j} \right)=\delta_{ij}$.
For $k+1$, let $\omega$ be a differential form of order $k+1$, then $\omega^L:X(\Omega) \times \cdots \times X(\Omega) \to C^\infty(\Omega)$ given by $\omega^L(L_1,\dots,L_k)=\omega(L_1,\dots,L_k,L)$ is a differential form of order $k$, then \begin{eqnarray} \omega(L_1,\dots,L_k,L)&=&\omega^L(L_1,\dots,L_k)\\ &=& \sum_{|J|=k}'\omega_{J}^{L} dx_J(L_1,\dots,L_k)\\ &=& \sum_{|J|=k}'\omega^{L}\left(\frac{\partial}{\partial x_{j_1}},\dots, \frac{\partial}{\partial x_{j_k}} \right) dx_J (L_1,\dots,L_k)\\ &=& \sum_{|J|=k}'\omega\left(\frac{\partial}{\partial x_{j_1}},\dots, \frac{\partial}{\partial x_{j_k}},L \right) dx_J (L_1,\dots,L_k)\\ &=& \sum_{|J|=k}'\omega\left(\frac{\partial}{\partial x_{j_1}},\dots, \frac{\partial}{\partial x_{j_k}},\sum_{j_{k+1}=1}^{N}a_{j_{k+1}}\frac{\partial}{\partial x_{j_{k+1}}} \right) dx_J (L_1,\dots,L_k)\\ &=& \sum_{|J|=k}'\sum_{j_{k+1}=1}^{N}a_{j_{k+1}}\omega\left(\frac{\partial}{\partial x_{j_1}},\dots, \frac{\partial}{\partial x_{j_k}},\frac{\partial}{\partial x_{j_{k+1}}} \right) dx_J (L_1,\dots,L_k)\\ &\vdots& \end{eqnarray} I would like to know how to continue the proof from here.