The product of two closed forms is closed

Question

A form $\beta^{p}$ is closed if $d\beta=0$.

A form $\beta^{p}$ is exact if $\beta^{p}=d\alpha^{p-1}$, for some form $\alpha^{p-1}$.

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How do you prove that the product of two closed forms is closed?

Answer 1

Remember that $\Bbb d (\alpha \wedge \beta) = (\Bbb d \alpha) \wedge \beta + (-1)^{|\alpha|} \alpha \wedge (\Bbb d \beta)$, where $|\alpha|$ is the degree of $\alpha$. Now, if $\Bbb d \alpha = 0$ and $\Bbb d \beta = 0$, the result follows easily.