I am working on the following exercise:
The function $f$ is called an integrating factor for the 1-form $\omega$ if $f({\bf x}) \neq 0$ for all $\bf x$ and $d(f\omega) = 0$. If the 1-form $\omega$ has an integrating factor, show that $\omega \wedge d\omega = 0$.
I am stuck here... I got $$d(f\omega) = df \wedge \omega + f \wedge d\omega = df \wedge \omega + f\ d\omega = 0$$ but that doesn't seem to get me anywhere. I also tried expanding this further (using the definition of $df$), but this gets quite ugly soon and didn't help either. The same goes for $\omega \wedge d\omega$.