Let $\dim(V) = n \geq 2$. Show that every alternating multilinear $n-1$ form is decomposable, that is it can be written as $\omega = \alpha \wedge \beta$ with $\alpha \in \Lambda^1V^*$, $\beta \in \Lambda^{n-2}V^*$.
My idea so far is to choose a basis for $V$ and $V^*$ and write $\omega = \sum_{j = 1}^n \omega_j e^1 \wedge \dots \wedge \hat{e^j}\wedge \dots \wedge e^n$.
If any of the $\omega_j = 0$, w. l. o. g. $j = n$, we have $\omega = \left( \sum_{j = 1}^{n-1} \omega_j e^1 \wedge \dots \wedge \hat{e^j}\wedge \dots \wedge e^{n-1} \right) \wedge e^n$ and we are done.
If not, maybe we can use the fact that we can divide through any of the $\omega_j$ and somehow define a new basis in which we will land in the first case?
PS: I'm sure that this question must have been asked before, as it is a common exercise, but I couldn't find it through the search function.
I think I managed to find an answer myself!
We define a new basis $(d^1, \dots, d^n)$ of $V^*$ by setting $d^i := e^i$ if $i \neq n$, $d^n := e^n + (-1)^{n-1} \frac{ \omega_n }{ \omega_1} e^1$.
Then (if my calculation was right) $$ \omega = \left( \sum_{j = 1}^{n-1} \omega_j d^1 \wedge \dots \wedge \hat{d^j}\wedge \dots \wedge d^{n-1} \right) \wedge d^n $$.