I'm going over the exterior covariant derivative $$d^\nabla : \Omega^k(E) \to \Omega^{k+1}(E)$$ of a vector bundle $E \to M$ and a connection $\nabla$ on $E$.
There are some peculiar notions which I'm facing. I'm trying to verify that $$d^\nabla( d^\nabla \omega) = F^\nabla \wedge \omega$$ where $F^\nabla$ is the curvature of $\nabla$.
The first thing I'm wondering is that can we write $\omega \in \Omega^k(E)$ as $\omega = \alpha \otimes s$ for $\alpha \in \Omega^k(M)$ and $s \in \Omega^0(E)$ always or do we need to assume locality or something?
If so, then
$$ \begin{align*} d^\nabla( d^\nabla \omega) &= d^\nabla( d^\nabla \alpha \otimes s) \\ &= d^\nabla( d\alpha \otimes s +(-1)^k\alpha \wedge \nabla s) \\ &= d^\nabla(d\alpha \otimes s) + (-1)^kd^\nabla(\alpha \wedge \nabla s) \end{align*} $$
but I'm not sure how to proceed from here. I also saw a note that the pairing that the pairing $F^\nabla \wedge \omega$ combines the wedge product $\Lambda^2 M \otimes \Lambda^k M \to \Lambda^{k+2}M$ with the evaluation map $\text{End}(E) \otimes E \to E$ which I did not really understand.