Let $M$ be a $n$ dimensional compact differentiable manifold. I would like to show that any exact differential form of degree $n$ vanishes at at least one point.
I think it is a generalization of the following fact : if $f$ is a diffferentiable function on $M$ then it either has a maximum or is constant so it's differential vanishes at at least one point.
edit I'm looking for a direct answer if possible maybe in the spirit of the above remark.
Say $\alpha$ is your exact degree $n$ form. Since it is exact, there exists $\tau$ such that $\alpha = d\tau$. Suppose now $M$ is orientable. Then, since it's also compact, we can do integration, in which case Stokes' Theorem tells us that $\int_M\alpha = \int_Md\tau = \int_{\partial M}\tau = 0$ since the boundary $\partial M$ of $M$ is empty. From this we know $\alpha$ must vanish at some point because otherwise it would be a volume form on $M$ which can't have $0$ volume.
If instead $M$ is non-orientable, take its orientation double cover $\pi: \widetilde{M} \rightarrow M$ (which you can check will still be compact) and pullback $\alpha$. Then $\pi^*\alpha$ will still be exact since $\alpha$ is, so we can repeat the argument upstairs to find that $\pi^*\alpha$ vanishes at some point $p \in \widetilde{M}$. Then you can find that $\alpha$ vanishes at $\pi(p) \in M$.
Edit: as pointed out by Georges in the comments, if we start out assuming (for sake of contradiction) that $\alpha$ is non-vanishing, then it would be an orientation form for $M$ and the argument in the first paragraph then suffices for a shorter proof.