TITLE: If $\omega^k$ is exact, is $\omega$ exact?
QUESTION [6 upvotes]: Let $\omega$ be a differential form on a manifold and suppose $\omega^k$ is exact (but non-zero). Is $\omega$ necessarily exact?

Note, the converse is true. If $\omega$ is exact, then $\omega^k$ is exact. To see this, let $\omega = d\alpha$, then $$\omega^k = (d\alpha)^k = d\alpha\wedge\dots\wedge d\alpha = d(\alpha\wedge d\alpha \wedge\dots\wedge d\alpha).$$
The case I am interested in, $\omega$ is a two-form and the manifold is compact, but I'd be interested to see what can be said in general.

REPLY [2 votes]: Consider the $2$-form 
$$\omega = dx\wedge dy + \sin (2\pi x)\, dz\wedge dw$$
on the $4$-torus $\mathbb R^4/\mathbb Z^4$. Since 
$$d\omega = 2\pi  \cos (2\pi  x)\, dx\wedge dz\wedge dw \ne 0,$$
the form $\omega$ is not exact. On the other hand,
$$
\omega\wedge\omega =  2 \sin (2\pi x)\,dx\wedge dy\wedge dz\wedge dw  = d (-\pi^{-1}\cos(2\pi x)\, dy\wedge dz\wedge dw) 
$$