= Solution
Use the <covariant exterior derivative> on the <endomorphism bundle> and remember that $F$ has degree two. The <Bianchi identity> is $d_AF=dF+A\wedge F-F\wedge A=0$. The local formula from part (b) gives
$$
\begin{aligned}
dF&=d(dA+A\wedge A)=dA\wedge A-A\wedge dA,\\
A\wedge F-F\wedge A
&=A\wedge dA+A\wedge A\wedge A-dA\wedge A-A\wedge A\wedge A\\
&=A\wedge dA-dA\wedge A.
\end{aligned}
$$
The two lines cancel, establishing
$$
\boxed{d_AF=0.}
$$
Because this is an <endomorphism>-valued equation and the operators transform covariantly under a change of <local frame>, this local calculation proves the global identity.
For a <line bundle> over $k=\mathbb R$ or $\mathbb C$, every fibre <endomorphism> is multiplication by a scalar. Thus $\operatorname{End}E$ is canonically the trivial scalar bundle, even when $E$ itself is nontrivial. In a <local frame> $A$ is a scalar one-form, so $A\wedge A=0$ and $F=dA$ locally. Also $A\wedge F=F\wedge A$, since their degrees are one and two. The <Bianchi identity> therefore reduces to the ordinary closedness condition
$$
\boxed{dF=0.}
$$
This does not assert that the local scalar one-forms $A$ combine into a global one-form.
For two <connections on a vector bundle> $\nabla,\nabla'$, the difference is $C^\infty(M)$-linear in the section: the two $df\otimes s$ terms cancel. It is consequently a global <endomorphism>-valued one-form $a$. In rank one, $a$ is a global ordinary scalar one-form. Locally $A'=A+a$, and the scalar cross terms and $a\wedge a$ vanish, giving
$$
F_{A'}-F_A=d(A+a)-dA=da.
$$
This is a global equality, so the two <closed differential forms> differ by an <exact differential form>. It proves the <connection-independent curvature class of a line bundle>:
$$
\boxed{[F_{A'}]=[F_A]\in H^2_{\mathrm{dR}}(M;k).}
$$
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