Exact top forms are not volume elements

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Let $M$ be a compact, orientable, smooth manifold with empty border. Prove that a volume element $\omega$ of $M$ cannot be an exact form (i.e., there is no $\eta\in\Omega^{n-1}(M)$ such that $\omega=\text{d}\eta$).

The first idea I had was to use Stoke's theorem. If there is an $\eta$ with $\omega=\text{d}\eta$, then we have $\int_{M}\omega=\int_{M}\text{d}\eta = \int_{\partial M=\emptyset}\eta=0$. Then I would argue that this is a contradiction, since $\omega$ is a volume element, which implies $\int_{M}\omega\neq0$.

I'm not sure if this argumentation is correct, because I still have some elementary questions: first, does it make sense to use Stoke's theorem when $\partial M$ is trivial? Second: is it true that $\int_{M}\omega\neq0$? How do I prove that?

Thanks!