= Solution
Fix the input-index convention
$$
De_i=\sum_j\theta_{ij}\otimes e_j,\qquad D^2e_i=\sum_j\Theta_{ij}\otimes e_j.
$$
Extend the <connection on a vector bundle> by the graded <Leibniz rule>, $D(\alpha\otimes s)=d\alpha\otimes s+(-1)^{\deg\alpha}\alpha\wedge Ds$. Applying it twice gives
$$
D^2e_i=\sum_jd\theta_{ij}\otimes e_j-\sum_{j,k}\theta_{ij}\wedge\theta_{jk}\otimes e_k,
$$
so
$$
\boxed{\Theta=d\theta-\theta\wedge\theta.}
$$
This is the <Cartan curvature equation with input indices>. In the more usual coefficient-column convention $A=\theta^t$, and $(A\wedge A)^t=-\theta\wedge\theta$ because the entries have degree one. Thus $F=dA+A\wedge A=\Theta^t$; the sign difference is entirely conventional.
The <Chern connection> of a <Hermitian vector bundle> with holomorphic structure is characterized by <metric compatibility> and $D^{0,1}=\bar\partial_E$. To construct it, use a <holomorphic local frame> and let $h_{ij}=\langle e_i,e_j\rangle$, with the <Hermitian inner product> linear in the first variable. The second defining property makes $\theta$ type $(1,0)$. The first gives
$$
dh=\theta h+h\overline\theta^t,\qquad\partial h=\theta h,
$$
and therefore the only possible matrix is
$$
\boxed{\theta=(\partial h)h^{-1}.}
$$
Conversely, this formula has type $(1,0)$ and satisfies both type components of the metric identity, because $h=\overline h^t$. Under a holomorphic frame change $e'=ge$, $h'=gh\overline g^t$ and direct differentiation gives
$$
\theta'=\partial g\,g^{-1}+g\theta g^{-1}.
$$
This is precisely the transformation obtained by differentiating $e'=ge$. The locally constructed connections glue, proving existence, and the forced formula proves uniqueness of the <Chern connection>.
The derivative of $h^{-1}$ gives $\partial\theta=\theta\wedge\theta$. Consequently the <Chern curvature> is
$$
\boxed{\Theta=\bar\partial\theta\in A^{1,1}(\operatorname{End}E),\qquad\bar\partial_{\operatorname{End}E}\Theta=0.}
$$
The second assertion is $\bar\partial^2\theta=0$ in a <holomorphic local frame>; holomorphic transition functions make it intrinsic. Thus the <curvature Dolbeault class> is $\Psi=[\Theta]\in H^1(M,\Omega_M^1\otimes\operatorname{End}E)$ by the <Dolbeault theorem>. Here $\Omega^1(\operatorname{End}E)$ denotes this tensor product. We use the identification with bundle-valued Dolbeault forms in which the antiholomorphic exterior factor is placed first, so $\bar\partial\theta$ has the stated positive descent sign.
For the stated transitions $e^{(\beta)}=g_{\alpha\beta}e^{(\alpha)}$, differentiating the frame relation yields
$$
\theta_\beta=g_{\alpha\beta}\theta_\alpha g_{\alpha\beta}^{-1}+(\partial g_{\alpha\beta})g_{\alpha\beta}^{-1}.
$$
Transport the alpha matrix to the beta frame before subtracting it. The difference is the holomorphic section
$$
\boxed{\sigma_{\alpha\beta}=(\partial g_{\alpha\beta})g_{\alpha\beta}^{-1}.}
$$
On a triple intersection, $g_{\alpha\gamma}=g_{\beta\gamma}g_{\alpha\beta}$, so
$$
\sigma_{\alpha\gamma}=\sigma_{\beta\gamma}+g_{\beta\gamma}\sigma_{\alpha\beta}g_{\beta\gamma}^{-1}.
$$
This is exactly the <Čech cocycle condition> after transporting all representatives to the gamma frame.
To identify its class, let $D_\alpha$ be the local holomorphic connection making the alpha frame parallel. Then $t_\alpha=D-D_\alpha$ is a genuine local section of $A^{1,0}(\operatorname{End}E)$, represented by $\theta_\alpha$, with $\bar\partial t_\alpha=\Theta$. Its <Čech coboundary>, using $t_\beta-t_\alpha$, is the intrinsic section represented by $\sigma_{\alpha\beta}$ in the beta frame. This local-primitive construction is the Čech–Dolbeault identification, and proves that $(\sigma_{\alpha\beta})$ maps to $\Psi$. For an arbitrary trivializing cover use its natural map to <sheaf cohomology>; a Stein refinement gives the usual Čech computation. The resulting object is the <Čech cocycle for the curvature Dolbeault class>.
If the metric changes, the difference $B=D'-D$ is a global smooth $(1,0)$ form with values in $\operatorname{End}E$. The local curvature formulas give $\Theta'-\Theta=\bar\partial B$, so $\Psi$ is metric independent. If holomorphic frames change by $e'_\alpha=u_\alpha e_\alpha$, the corresponding local tensors $t'_\alpha-t_\alpha$, expressed in the old frame, are $B_\alpha=u_\alpha^{-1}\partial u_\alpha$. They are holomorphic, and their differences change the cocycle by $\delta B$. Hence \b[$\Psi$ depends only on the holomorphic bundle, not on its metric or trivializations].
Apply <exterior product> and endomorphism composition to curvature to define
$$
\boxed{\Psi^{(k)}=[\Theta^{\wedge k}]\in H^k(M,\Omega_M^k\otimes\operatorname{End}E),\qquad\operatorname{tr}\Psi^{(k)}\in H^k(M,\Omega_M^k).}
$$
These are closed for the <Dolbeault operator> by the product rule. Their independence can also be verified without commuting matrices. If $\Theta_1-\Theta_0=\bar\partial B$, then
$$
\Theta_1^k-\Theta_0^k=\bar\partial\left(\sum_{j=0}^{k-1}\Theta_1^j\wedge B\wedge\Theta_0^{k-1-j}\right).
$$
This follows by the noncommutative telescoping identity and the even total degree of curvature. The matrix products are interpreted in the fixed frame convention; powers of the degree-two curvature transpose to the corresponding usual curvature powers. The trace contraction is invariant under frame changes, so its class is well defined. This proves <metric independence of curvature Dolbeault powers>; for $k>\dim M$ the forms and groups in question vanish.
Finally take a smooth unitary frame. Metric compatibility gives $\theta^\dagger=-\theta$, and the <Cartan curvature equation with input indices> gives $\Theta^\dagger=-\Theta$. Thus
$$
\overline{\operatorname{tr}(\Theta^k)}=(-1)^k\operatorname{tr}(\Theta^k),\qquad \overline{\left(\frac{i}{2\pi}\right)^k\operatorname{tr}(\Theta^k)}=\left(\frac{i}{2\pi}\right)^k\operatorname{tr}(\Theta^k).
$$
The second expression is therefore a real differential form. It is also $d$-closed: differentiating Cartan's equation gives $d\Theta=\theta\wedge\Theta-\Theta\wedge\theta$, and cyclicity of trace cancels the resulting commutator in $d\operatorname{tr}(\Theta^k)$. On a compact <Kähler manifold>, the <Hodge decomposition theorem for compact Kähler manifolds> identifies $H^k(M,\Omega_M^k)$ with the $(k,k)$ component of complex <de Rham cohomology>. Hence
$$
\boxed{\left(\frac{i}{2\pi}\right)^k\operatorname{tr}(\Psi^{(k)})\text{ determines a class in }H^{2k}_{\mathrm{DR}}(M,\mathbb R).}
$$
It is $k!$ times the degree-$2k$ component of the <Chern character>. Reality does not assert integrality of these individual power-sum classes.
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