Given a nowhere vanishing vector field, say $E_1$, on a manifold $M^n$, it should be possible to extend this locally to a basis of vector fields $E_1,\ldots , E_n$ so that $E_i = \frac{\partial}{\partial x_i}$ for some coordinates $x_i$. (For example, locally we can choose a chart where $E_1$ looks like $\frac{\partial}{\partial x_1}$ and then just define $E_i = \frac{\partial}{\partial x_i}$).
For such a frame there we may define a co-frame on $T^*M$, $\varphi^i$ so that $\varphi^i(E_j) = \delta_{ij}$. By the intrinsic definition of the exterior derivative, \begin{equation} d\varphi^1(E_i,E_j) = E_i(\varphi^1(E_i)) - E_j(\varphi^1(E_j) - \varphi^1([E_i,E_j]) . \end{equation} The first two terms will be zero as they are the derivative of either the constant function 0 or 1. Similarly, $[E_i,E_j] = [\frac{\partial}{\partial x_i},\frac{\partial}{\partial x_j}] \equiv 0$.
Finally, since $d\varphi^1$ is a two form, we see by linearity that $d\varphi^1 \equiv 0$ .
In fact, any nonvanishing 1-form $\omega$ can (locally) be expressed as the dual of a nonvanishing vector in some coordinate framing. This would seem to imply that any nowhere zero 1-form has zero exterior derivative.
Since I do not believe this is true, there must be some flaw to my reasoning, but I cannot see it.