 Solution

An orientation of a smooth $n$-dimensional manifold can be specified by a nowhere-vanishing smooth $n$-form $\omega$: an ordered tangent basis is positive precisely when $\omega$ evaluates positively on it. Multiplication by a positive smooth function preserves the orientation. This is the differential-form formulation of the equivalent formulations of orientability of a smooth manifold.

Use the standard convention that the compact connected manifold is nonempty and has no boundary. Choose a positive volume form $\omega$ for its orientation. It is a closed differential form, because there are no nonzero $(n+1)$-forms. For $n>0$, compactness makes its integral finite, and positivity on a coordinate neighborhood together with nonnegativity elsewhere givesIf $\omega =d\eta$, the Generalized Stokes theorem would give ${\int }_M\omega ={\int }_{\partial M}\eta =0$, a contradiction. ThereforeFor $n=0$, a nonempty connected manifold is a point, and $H_{\mathrm{dR}}^0(M)=\mathbb{R}$ directly. The boundary hypothesis matters: a closed interval is an oriented compact connected manifold with boundary with $H_{\mathrm{dR}}^1=0$. Thus the result is about ordinary boundaryless manifolds, not an unrestricted version including boundary.