I don't know what to do to prove the following statement:
Let $U \subset \mathbb R^n$ be an open set and let $\alpha$ be a $k$-form on $U$ and $\beta$ be an $l$-form on $U$. Suppose both $\alpha, \beta$ are closed forms and that $\beta$ is exact. Then $\alpha \wedge \beta$ is exact.
Every assistance would be very much appreciated.
If $b$ satisfies $db=\beta$, and $\alpha$ is a closed $k$-form, then $$d\omega=\alpha\wedge\beta$$ where $\omega=(-1)^k\alpha\wedge b$. The proof works for forms on any manifold.