= Solution
A <connection on a vector bundle> $E\to M$ is a linear map $D:\Gamma(E)\to\Omega^1(M;E)$ with $D(fs)=df\otimes s+fDs$. Write a <local frame> as a row $e=(e_1,\ldots,e_r)$, and a section as $s=eu$ with a coefficient column $u$. If the <connection one-form> is the <matrix> $A=\sum_iA_i\,dx^i$, then
$$
\boxed{D(eu)=e(du+Au),\qquad
(D_is)^a=\partial_i u^a+\sum_b(A_i)^a{}_bu^b.}
$$
This fixes the <matrix> convention for all subsequent signs.
For an $E$-valued <differential form> of degree $k$, represented locally by a column $\alpha$ of ordinary forms, its <covariant exterior derivative> is
$$
\boxed{d_A\alpha=d\alpha+A\wedge\alpha.}
$$
The defining <graded Leibniz rule> with a scalar form $\omega$ of degree $j$ is
$$
d_A(\omega\wedge\alpha)=d\omega\wedge\alpha+(-1)^j\omega\wedge d_A\alpha.
$$
It follows by moving the one-form $A$ past $\omega$, which introduces $(-1)^j$. It uniquely extends the connection on sections.
The induced <endomorphism bundle connection> is characterized on an endomorphism section $T$ by $(D^{\rm End}T)s=D(Ts)-T(Ds)$. For a matrix-valued form $\beta$ of degree $k$, the extension is
$$
\boxed{d_A^{\rm End}\beta=d\beta+A\wedge\beta-(-1)^k\beta\wedge A.}
$$
Indeed it is exactly the formula that makes
$$
d_A(\beta\wedge\alpha)=(d_A^{\rm End}\beta)\wedge\alpha+(-1)^k\beta\wedge d_A\alpha.
$$
The <endomorphism-valued exterior product> uses <matrix> composition as well as the ordinary <exterior product>, so factors cannot in general be interchanged.
Define the <curvature form of a connection> by $D^2$ acting on sections, or equivalently by
$$
F(X,Y)s=D_XD_Ys-D_YD_Xs-D_{[X,Y]}s.
$$
It is tensorial in both vector arguments and in $s$: the derivatives of multiplying functions cancel by the connection rule. For example $D^2(fs)=D(df\,s+fDs)=fD^2s$, since the two $df\wedge Ds$ terms cancel and $d^2f=0$. Thus curvature is a two-form with values in the <endomorphism bundle>. Locally, applying the graded rule gives
$$
(d+A\wedge)^2\alpha=(dA+A\wedge A)\wedge\alpha,
$$
so
$$
\boxed{F(A)=dA+A\wedge A,\qquad
F_{ij}=\partial_iA_j-\partial_jA_i+[A_i,A_j].}
$$
For a <change of frame of a vector-bundle connection> $e'=eg$, the coefficients satisfy $u'=g^{-1}u$, and differentiation gives
$$
A'=g^{-1}Ag+g^{-1}dg,\qquad d_{A'}=g^{-1}d_Ag.
$$
Squaring yields $F'=g^{-1}Fg$, precisely the transformation rule of an endomorphism-valued two-form. This independently proves that the local expressions define one global <curvature form of a connection>.
For the <Bianchi identity>, use the endomorphism formula with degree two:
$$
\begin{aligned}
d_A^{\rm End}F
&=dF+A\wedge F-F\wedge A\\
&=(dA\wedge A-A\wedge dA)
+(A\wedge dA+A\wedge A\wedge A)\\
&\hspace{1em}-(dA\wedge A+A\wedge A\wedge A)=0.
\end{aligned}
$$
Every term cancels with its indicated sign. Therefore
$$
\boxed{d_A F(A)=0,}
$$
where $d_A$ here means the induced derivative on the <endomorphism bundle>.
For a <line bundle>, the <endomorphism bundle> is canonically the trivial scalar bundle: every fibre endomorphism is a scalar times the identity. Scalar multiplication commutes, so $A\wedge A=0$ and the endomorphism covariant derivative on scalar-valued forms reduces to $d$. The <Bianchi identity> consequently says $dF=0$.
If $D'$ is another connection, their difference is $C^\infty$-linear in sections because the Leibniz terms cancel. Hence $D'-D=\eta$ is a global endomorphism-valued one-form; for a <line bundle> it is an ordinary global scalar one-form. Locally $A'=A+\eta$, so
$$
F(D')-F(D)=d\eta.
$$
Thus the two closed curvatures differ by an exact form, proving the <connection-independent curvature class of a line bundle>:
$$
\boxed{[F(D')]=[F(D)]\in H^2_{\rm dR}(M;\mathbb K).}
$$
No global trivialization of the <line bundle> itself was assumed; it is its <endomorphism bundle> that has the canonical scalar identification.
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