Consider a $C^1$ differential $k$-form $\omega$ on an open set $U \subset \mathbb R^n$. A non-vanishing $C^1$ function $f: \mathbb R^n \to \mathbb R$ is called an integrating factor for $\omega$ if $d(f\omega) = 0$. If $k$ is odd and an integrating factor exists, I want to show that $\omega \land d\omega = 0 $.
Here's my proof:
$$d(f\omega \land \omega) = d(f\omega) \land \omega + (-1)^k f\omega \land d\omega = 0 \land \omega + (-1)^k f\omega \land d\omega = (-1)^k f(\omega \land d\omega)$$
And $f\omega \land \omega = f(\omega \land \omega) = 0$, so $d(f\omega \land \omega) = 0$ and hence $(-1)^k f(\omega \land d\omega) = 0$. If $f$ never vanishes, then $(-1)^k f$ never vanishes, so that $\omega \land d\omega = 0$.
Clearly, my proof doesn't use that $k$ must be odd. What did I do wrong here?
$\omega \wedge \omega=0$ is only true for odd forms.
Observe $$(dx_1\wedge dx_2 +dx_3 \wedge dx_4) \wedge (dx_1\wedge dx_2 +dx_3 \wedge dx_4)=2 dx_1 \wedge dx_2 \wedge dx_3 \wedge dx_4 \neq 0.$$