I'm trying to prove the following statement, and I think I'm nearly there but I can't seem to figure out the last step. Any tips/feedback/hints would be greatly appreciated.
A smooth $n$-manifold $M$ admits an orientation if and only if $M$ has a nowhere vanishing $n$-form.
Let $\omega$ be such a form. In a chart $(U,\varphi)$, there is a map $f:\varphi_\alpha(U)\to\mathbb{R}$ so that $\omega = f d x^1_\alpha\wedge\dots \wedge d x^n_\alpha$. In particular $f=\omega(\partial_1^\alpha,\dots, \partial_n^\alpha)$. Now choose a chart so that $f(x)>0$, and let $(U,\varphi_\alpha)$ be another such chart with $g:\varphi_\beta(U_\beta)\to\mathbb{R}$ that has $\omega = g d x^1_\beta\wedge\dots d x^n_\beta$ and $g(x)>0$. Now, \begin{equation} g = \omega(\partial_1^\beta,\dots, \partial_n^\beta) = \omega (D_1 (\varphi_\alpha\circ\varphi_\beta^{-1})_1^j\partial_j^\beta,\dots, D_n (\varphi_\alpha\circ\varphi_\beta^{-1})_n^j\partial_j^\alpha) \end{equation} But this is just $\det(D \varphi_\alpha\circ\varphi_\alpha^{-1})f$. Building an atlas this way gives an orientation. Now suppose we have an oriented atlas $\{(U_\alpha,\varphi_\alpha)\}$, and write, in each chart $\omega_\alpha = d x^1_\alpha\wedge\dots\wedge d x^n_\alpha$, and choose a partition of unity $\{\rho_\alpha\}$ subbordinate to $\{(U_\alpha,\varphi)\}$. Define $\omega =\sum_\alpha \omega_\alpha \rho_\alpha$. Fix $p\in M$. For open $U\subset M$ with $p\in U$, we know $\sum_\alpha \rho_\alpha \omega_\alpha$ is a finite sum $\sum_{i=1}^k \rho_i \omega_i$. Then, around $p\in U_i$ \begin{equation} \omega(\partial^i_1,\dots,\partial_n^i) = \sum_{i=1}^k \rho_i \omega_i (\partial^i_1,\dots,\partial_n^i) \end{equation}