[Math] For $(n-1)$-form $\omega$ on $M^{n}$ compact, orientable without boundary, then $d\omega$ vanish for some point

Question

Let $M^{n}$ manifold compact, orientable without boundary and $\omega$ $(n-1)$-form then there is $p\in M$ such that $d\omega(p)=0$.

This is for my homework of integration on manifolds & Stokes theorem and I really don't know how use it.
A hint please

Answer 1

I think the point here is that an n-form which is non-zero everywhere defines a volume form and can be integrated to give a volume for the manifold which is positive. If you integrate an exact form you get zero, since there is no boundary. Hence the exact form cannot be a volume form.