[proofplan]
We descend the standard coordinate frame of $T^{1,0}\mathbb{C}^n$ to the quotient $X=\mathbb{C}^n/\Lambda$. The key point is that every deck transformation is a translation, and the differential of a translation is the identity, so the coordinate vector fields are $\Lambda$-invariant. Their descended vector fields form a global holomorphic frame of $T^{1,0}X$, which yields a holomorphic bundle isomorphism. The vanishing of all positive Chern classes then follows from the normalization property of Chern classes for product complex vector bundles.
[/proofplan]
[step:Descend the standard coordinate vector fields through the quotient map]
Let
\begin{align*}
\pi:\mathbb{C}^n \to X=\mathbb{C}^n/\Lambda
\end{align*}
be the quotient map. For each $\lambda \in \Lambda$, define the deck transformation
\begin{align*}
\tau_\lambda:\mathbb{C}^n \to \mathbb{C}^n
\end{align*}
by $\tau_\lambda(z)=z+\lambda$. Since $\tau_\lambda$ is a holomorphic translation, its complex differential at every point $z \in \mathbb{C}^n$ is the identity complex-[linear map](/page/Linear%20Map)
\begin{align*}
d(\tau_\lambda)_z:T_z^{1,0}\mathbb{C}^n \to T_{z+\lambda}^{1,0}\mathbb{C}^n.
\end{align*}
For $j \in \{1,\dots,n\}$, let
\begin{align*}
E_j:\mathbb{C}^n \to T^{1,0}\mathbb{C}^n
\end{align*}
denote the standard holomorphic coordinate vector field $E_j(z)=\partial_{z_j}|_z$. The identity differential of translations gives
\begin{align*}
d(\tau_\lambda)_z(E_j(z))=E_j(z+\lambda).
\end{align*}
Because $\pi \circ \tau_\lambda=\pi$, differentiating gives
\begin{align*}
d\pi_{z+\lambda}\circ d(\tau_\lambda)_z=d\pi_z.
\end{align*}
Therefore
\begin{align*}
d\pi_{z+\lambda}(E_j(z+\lambda))=d\pi_z(E_j(z)).
\end{align*}
Thus the formula
\begin{align*}
V_j(\pi(z)):=d\pi_z(E_j(z))
\end{align*}
defines a well-defined section
\begin{align*}
V_j:X \to T^{1,0}X.
\end{align*}
[guided]
The quotient map $\pi:\mathbb{C}^n \to X$ identifies two points $z,z' \in \mathbb{C}^n$ exactly when $z'=z+\lambda$ for some $\lambda \in \Lambda$. To define a vector field on $X$ by pushing forward a vector field from $\mathbb{C}^n$, we must check that the answer does not depend on which representative $z$ of the point $\pi(z)$ we choose.
For each lattice element $\lambda \in \Lambda$, the corresponding deck transformation is the holomorphic map
\begin{align*}
\tau_\lambda:\mathbb{C}^n \to \mathbb{C}^n
\end{align*}
given by $\tau_\lambda(z)=z+\lambda$. Its differential is the identity on tangent vectors, viewed between the tangent spaces at $z$ and $z+\lambda$:
\begin{align*}
d(\tau_\lambda)_z:T_z^{1,0}\mathbb{C}^n \to T_{z+\lambda}^{1,0}\mathbb{C}^n.
\end{align*}
Hence the standard coordinate vector field
\begin{align*}
E_j:\mathbb{C}^n \to T^{1,0}\mathbb{C}^n
\end{align*}
defined by $E_j(z)=\partial_{z_j}|_z$ satisfies
\begin{align*}
d(\tau_\lambda)_z(E_j(z))=E_j(z+\lambda).
\end{align*}
The quotient map is constant along $\Lambda$-orbits, so $\pi \circ \tau_\lambda=\pi$. Differentiating this identity at $z$ gives
\begin{align*}
d\pi_{z+\lambda}\circ d(\tau_\lambda)_z=d\pi_z.
\end{align*}
Applying both sides to $E_j(z)$ yields
\begin{align*}
d\pi_{z+\lambda}(E_j(z+\lambda))=d\pi_z(E_j(z)).
\end{align*}
This is exactly the well-definedness condition. If $z$ and $z+\lambda$ represent the same point of $X$, then pushing forward the coordinate vector field from either representative gives the same tangent vector in $T_{\pi(z)}^{1,0}X$. Therefore the formula
\begin{align*}
V_j(\pi(z)):=d\pi_z(E_j(z))
\end{align*}
defines a genuine global section $V_j:X \to T^{1,0}X$.
[/guided]
[/step]
[step:Show that the descended vector fields are holomorphic and form a frame]
Fix $x \in X$ and choose $z \in \mathbb{C}^n$ with $\pi(z)=x$. Since $\Lambda$ is discrete, the quotient map $\pi$ is a local biholomorphism. Hence there is an open neighbourhood $U \subset X$ of $x$ and a holomorphic section
\begin{align*}
s:U \to \mathbb{C}^n
\end{align*}
of $\pi$ over $U$, meaning $\pi \circ s=\operatorname{id}_U$.
On $U$, the vector field $V_j$ is given by the holomorphic expression
\begin{align*}
V_j(y)=d\pi_{s(y)}(E_j(s(y)))
\end{align*}
for $y \in U$. Since $s$, $E_j$, and $\pi$ are holomorphic, $V_j$ is holomorphic on $U$. As $x$ was arbitrary, $V_j$ is a global holomorphic vector field on $X$.
For each $z \in \mathbb{C}^n$, the complex-linear map
\begin{align*}
d\pi_z:T_z^{1,0}\mathbb{C}^n \to T_{\pi(z)}^{1,0}X
\end{align*}
is an isomorphism because $\pi$ is a local biholomorphism. Since $E_1(z),\dots,E_n(z)$ is a basis of $T_z^{1,0}\mathbb{C}^n$, the vectors
\begin{align*}
V_1(\pi(z)),\dots,V_n(\pi(z))
\end{align*}
form a basis of $T_{\pi(z)}^{1,0}X$. Thus $V_1,\dots,V_n$ form a global holomorphic frame of $T^{1,0}X$.
[/step]
[step:Build the holomorphic bundle isomorphism from the global frame]
Define the bundle map
\begin{align*}
\Phi:X \times \mathbb{C}^n \to T^{1,0}X
\end{align*}
by
\begin{align*}
\Phi(x,a_1,\dots,a_n)=\sum_{j=1}^n a_j V_j(x).
\end{align*}
This map covers the identity map on $X$, is complex-linear on each fibre, and is holomorphic because each $V_j$ is holomorphic.
For every $x \in X$, the vectors $V_1(x),\dots,V_n(x)$ form a basis of $T_x^{1,0}X$. Hence the fibre map
\begin{align*}
\Phi_x:\mathbb{C}^n \to T_x^{1,0}X
\end{align*}
is a complex-linear isomorphism for every $x \in X$. Therefore $\Phi$ is a holomorphic vector bundle isomorphism
\begin{align*}
X \times \mathbb{C}^n \cong T^{1,0}X.
\end{align*}
[/step]
[step:Deduce the vanishing of the positive Chern classes]
The holomorphic vector bundle $T^{1,0}X$ is holomorphically isomorphic to the product rank $n$ complex vector bundle $X \times \mathbb{C}^n$. Chern classes are invariant under complex vector bundle isomorphism, and the total Chern class of a product complex vector bundle is
\begin{align*}
c(X \times \mathbb{C}^n)=1.
\end{align*}
Thus
\begin{align*}
c(T^{1,0}X)=1.
\end{align*}
Equating homogeneous components gives
\begin{align*}
c_k(T^{1,0}X)=0
\end{align*}
for every integer $k \geq 1$. This proves the theorem.
[/step]
