Let $M$ be a connected, orientable $n$-manifold without boundary. A well known fact is that the top cohomology $H^n(M, \mathbb{R})$ vanishes if and only if $M$ is not compact, but I have not been able to find an intuitive proof.
One direction is simple. If $M$ is compact, then any volume form is closed but not exact—if it were exact, it would integrate to zero. This proves $M$ compact $\Rightarrow$ $H^n(M, \mathbb{R}) \neq 0.$
There are a few questions on math.SE about the reverse direction, but all of the answers invoke Poincare duality: for non-compact $M$, we have $H^n(M, \mathbb{R}) = H_0^{c}(M) = 0.$ I feel it should be possible to prove the desired statement without any knowledge of singular homology, and would like a proof along those lines.
So the question is this: why, on a non-compact orientable manifold, is every top form exact?