Every $(n-1)$-form is decomposable
Solution 1:
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) $$ \omega = \left( \sum_{j = 1}^{n-1} \omega_j d^1 \wedge \dots \wedge \hat{d^j}\wedge \dots \wedge d^{n-1} \right) \wedge d^n $$.