Let $V$ be a vector space of dimension $d<\infty$, and let $v\in V\setminus\{0\}$. Let $f\in \Lambda^p V$ be such that $v\wedge f=0$. I want to show that there exists a $g\in \Lambda^{p-1}V$ such that $f=v\wedge g$.
My attempt: We can find a basis $\{e_1,\dots, e_d\}$ starting from $v=:e_1$. Let the dual basis be defined by $\{\epsilon^1,\dots, \epsilon^d\}$. My guess is to take $$ g(\tau^1,\dots,\tau^{p-1}):=f(\epsilon^1, \tau^1,\dots,\tau^{p-1}), \quad \tau^i\in V^*. $$ Then clearly $g\in \Lambda^{p-1}V$, since swapping any two $\tau^i$s creates a sign change, thanks to the skew-symmetry of $f$. I tried showing that the action of $f$ and of $v\wedge g$ is the same, but couldn't yet manage to show this using $v\wedge f=0$. I would be grateful for a hint or outline. Thank you!
First, I'll fix the notation $v\in\Lambda^1V$, and $\mathcal{V}\in V$ for the vector corresponding to $v$ through a basis, inner product, or something else. This is characterized by $v(\mathcal{V}) = 1$.
You've sorta discovered the interior product of forms with vector fields and have defined $g = \iota_\mathcal{V}(f)$. Taking a look at the properties of the inner product at the previous hyperlink, we have
$$ \iota_\mathcal{V}(v\wedge f) = \iota_\mathcal{V}(v)\wedge f - v\wedge \iota_\mathcal{V}(f). $$
Now, the result will follow from $$ \iota_\mathcal{V}(v) \wedge f= 1\wedge f = f $$ and from the hypothesis $v\wedge f = 0$.