Let $\beta$ be a $k$-form. Show that $d(\beta \wedge d\beta)=0$ if $k$ is even.
I get that $d(\beta \wedge d\beta)=d\beta \wedge d \beta + (-1)^k\beta \wedge d^2\beta=d\beta \wedge d \beta$.
Why doesnt this just equal $0$? I thought that if $dx^{(i)} = f^{(i)}$ then $dx^{(i)} \wedge dx^{(i)}=0 \ \forall i$?
Expanding on John's comment: the wedge product on differential forms is graded commutative (or skew commutative), meaning that for $\alpha, \beta \in \Omega^*(M)$, you have: $$\beta \wedge \alpha = (-1)^{|\alpha| |\beta|} \alpha \wedge \beta,$$ where $|\alpha|$, $|\beta|$ are the respective degrees of $\alpha$ and $\beta$. In particular if $\omega$ is a $k$-form, then $\omega \wedge \omega = (-1)^{k^2} \omega \wedge \omega$.
If $k$ is odd this implies $\omega \wedge \omega = - \omega \wedge \omega \implies \omega \wedge \omega = 0$. On the other hand if $k$ is even you only have the trivial identity $\omega \wedge \omega = \omega \wedge \omega$, and this doesn't imply that this form is zero.
It is true that you always have $dx^{(i)} \wedge dx^{(i)} = 0$, but this is because $dx^{(i)}$ has degree $1$, which is odd.