= Solution
The <smooth exponential sequence> is
$$
0\longrightarrow\underline{\mathbb Z}\longrightarrow
\mathcal C^\infty(\mathbb C)
\xrightarrow{\exp(2\pi i\,\cdot)}
\mathcal C^\infty(\mathbb C^\times)\longrightarrow1.
$$
Local logarithms make it exact as a sequence of sheaves. A <line bundle> has a transition class in $H^1(M,\mathcal C^\infty(\mathbb C^\times))$; its image under the connecting map defines the <First Chern class>. For a holomorphic <line bundle> the holomorphic exponential sequence gives the same class by naturality.
Here is the explicit <Čech-de Rham curvature descent>. On a good cover, write $e_j=e_i g_{ij}$ and choose logarithms $g_{ij}=\exp(2\pi i f_{ij})$. The integer cocycle
$$
c_{ijk}=f_{ij}+f_{jk}-f_{ik}
$$
represents $c_1(E)$. By the proved frame formula,
$$
A_j-A_i=2\pi i\,df_{ij}.
$$
Put $B_i=-A_i/(2\pi i)$ and $F=i\Theta/(2\pi)=dB_i$. Then $\delta B=-df$ and $\delta f=c$. In the <Čech-de Rham double complex>, with total differential $D_{\mathrm{tot}}=\delta+(-1)^p d$ on Čech degree $p$,
$$
D_{\mathrm{tot}}B=F-df,\qquad
D_{\mathrm{tot}}f=c-df,\qquad
F-c=D_{\mathrm{tot}}(B-f).
$$
Thus the global <closed differential form> and the integer cocycle represent the same class under <de Rham theorem>:
$$
\boxed{\left[\frac{i}{2\pi}\Theta\right]=c_1(E)_{\mathbb C}.}
$$
This proves integrality, including the sign and normalization. It identifies the image of the integral class; <vector-bundle curvature> alone cannot recover torsion classes lost in passage to complex coefficients.
Two connections on the same <line bundle> differ by a global scalar one-form $a$. Their <vector-bundle curvatures> satisfy $\Theta_1-\Theta=da$, so the normalized representatives differ by an <exact differential form>. Hence the class is independent of the connection.
Use the integral normalization of the <Fubini-Study form>. On the chart $Z_j\ne0$, define the <integrally normalized Fubini-Study form> by
$$
\boxed{\omega_{\mathrm{int}}=\frac{i}{2\pi}\partial\bar\partial
\log\!\left(\frac{\sum_{k=0}^n|Z_k|^2}{|Z_j|^2}\right).}
$$
The potentials on overlaps differ by the logarithm of the squared modulus of a nowhere-zero holomorphic function, whose $\partial\bar\partial$ is zero. Thus the forms patch and are closed. The dual of the tautological metric on $\mathcal O(1)$ has local squared norm $h=(1+\sum|z_k|^2)^{-1}$. Its <Chern connection> has $A=\partial\log h$ and <vector-bundle curvature> $\Theta=\bar\partial\partial\log h=\partial\bar\partial\log(1+\sum|z_k|^2)$. Therefore \b[$[\omega_{\mathrm{int}}]=c_1(\mathcal O(1))$], proving the requested integrality. With the unnormalized convention $\omega_{FS}=i\partial\bar\partial\log(1+\sum|z_k|^2)$, this is $\omega_{\mathrm{int}}=\omega_{FS}/(2\pi)$.
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