Top deRham cohomology group of a compact orientable manifold is 1-dimensional

Question

Let $M$ be a compact smooth orientable manifold of dimension $n$. I am looking for a simple proof that $H_\mathrm{dR}^n(M) \cong \mathbb R$. Equivalently, an $n$-form which integrates to 0 is exact. I can show this via a rather indirect argument as follows: we know $H_\mathrm{dR}^n(M) \cong H^n(M, \mathbb R)$, where $H^n$ denotes the singular cohomology. By the universal coefficient theorem (and the fact that $\mathbb R$ is a field) this is isomorphic to $\operatorname{Hom}(H_n(M, \mathbb Z) , \mathbb R)$. From the (rather lengthy) proof in Section 3.3 of Hatcher's Algebraic Topology, we find that $H_n(M, \mathbb Z)$ is isomorphic to $\mathbb Z$, and so $\operatorname{Hom}(H_n(M, \mathbb Z) , \mathbb R) \cong \mathbb R$. However, it seems like there should be a simpler way to prove this. Does anyone know of one?