I see, you did not specify 1-forms. Let me type this in, perhaps you can decide the status of the argument for other $p$-forms.
This is Cartan's Lemma, which is about 1-forms. On a smooth manifold (or small open neighborhood in one) of dimension $n,$ let $p \leq n.$ Let $\omega_1, \ldots, \omega_p$ be 1-forms that are (pointwise) linearly independent. Then let t $\theta_1, \ldots, \theta_p$ be 1-forms such that $ \sum_{i=1}^p \theta_i \wedge \omega_i = 0. $ Then there are smooth functions $A_{ij} = A_{ji}$ such that $ \forall i \leq p, \; \; \; \theta_i = \sum_{j=1}^p A_{ij} \omega_j \; \; . $
So, extend your 1-form $\alpha$ to a full basis $\omega_1, \ldots, \omega_{n-1}, \omega_n = \alpha.$ You are saying all wedges are zero, so $ \sum_{i=1}^{n-1} \alpha \wedge \omega_i = 0. $ Then $ \alpha = \sum_{j=1}^{n-1} A_{ij} \omega_j \; \; . $ However, $\omega_n = \alpha$ is part of the basis, which is contradicted, so this contradicts the ability to form such a basis and the assumption that $\alpha \neq 0.$