$V$ is a finite dimensional vector space over a field $\mathbb{k}$ of characteristic 0. For a nonzero covector $\xi \in V^*$, I need to compute the cohomologies of the complexes $\Lambda V$ and $\Lambda V^*$ with the differentials:
$$\partial_\xi:\Lambda^mV \rightarrow \Lambda^{m-1}V,\ \ \eta \mapsto \partial_\xi\eta \ \ \ \ \ \& \ \ \ \ \ \xi:\Lambda^mV^* \rightarrow \Lambda^{m+1}V^*,\ \ \omega \mapsto \xi \wedge \omega$$
The operator $\partial_\xi$ acts on $\eta \in \Lambda^mV$ as follows: we write $\eta′ \in V^{\otimes(m − 1)}$ for the contraction in the first tensor factor between $\xi$ and the unique alternating tensor $\tilde{\eta} \in V^{\otimes m}$ mapped to $\eta$ under the projection $V^{\otimes m} \twoheadrightarrow \Lambda^mV$ . Then $\partial_\xi \eta$ is defined as the image of $\eta$ under the projection $V^{\otimes (m-1)} \twoheadrightarrow \Lambda^{m-1}V$ multiplied by $m = \deg \eta$ . In terms of dual bases $e_i$ and $\xi_i$ in $V$ and $V^*$ , $\partial_{\xi_i} = \frac{\partial}{\partial e_i}$
Any kind of hint(s) will be appreciated.
Suppose $\omega\in \Lambda^m V^*$ is in the kernel of $\xi$, i.e. $\xi\wedge \omega=0$. Take a vector $X$ such that $i_X\xi:=\xi(X)=1$. Then we have that $$ 0=i_X(\xi \wedge \omega)=i_X(\xi)\wedge\omega-\xi\wedge i_X(\omega)=\omega-\xi\wedge i_X(\omega) $$ Hence $\omega=\xi\wedge i_X(\omega)$. Thus if $\omega$ is in the kernel of $\xi$ it also lies in the image of $\xi$ (of one degree lower). It follows that the cohomology vanishes.