[proofplan]
Choose a Hermitian metric on $V$ and a compatible complex-linear connection $\nabla$. The induced conjugate connection on $\overline V$ has curvature matrix obtained by complex conjugating the curvature matrix of $\nabla$. In a unitary local frame, the curvature matrix of $\nabla$ is skew-Hermitian, so the Chern-Weil determinant for $\overline V$ is obtained from the determinant for $V$ by replacing the curvature variable by its negative. The homogeneous degree $k$ part of the determinant therefore gains the factor $(-1)^k$, and the Chern-Weil representative theorem identifies these forms with the corresponding de Rham Chern classes.
[/proofplan]
[step:Choose a unitary connection and form its conjugate connection]
Let $r\ge 0$ denote the complex rank of $V$. Since $M$ is paracompact, choose a smooth Hermitian metric $h$ on $V$, and choose an $h$-compatible complex-linear connection
\begin{align*}
\nabla:\Gamma(V)\to \Omega^1(M;V).
\end{align*}
Let $\overline V$ denote the same underlying real vector bundle as $V$, with complex scalar multiplication defined by
\begin{align*}
z\cdot_{\overline V} v:=\overline z\,v
\end{align*}
for $z\in\mathbb C$ and $v$ in the underlying real [vector space](/page/Vector%20Space) of a fiber of $V$.
Define the conjugate connection
\begin{align*}
\overline\nabla:\Gamma(\overline V)\to \Omega^1(M;\overline V)
\end{align*}
as follows. If $s\in\Gamma(V)$ is a smooth section and $\overline s\in\Gamma(\overline V)$ is the corresponding section of the conjugate bundle, then for every smooth vector field $X\in\mathfrak X(M)$ set
\begin{align*}
\overline\nabla_X\overline s:=\overline{\nabla_Xs}.
\end{align*}
This is a complex-linear connection on $\overline V$ because scalar multiplication in $\overline V$ conjugates the scalar multiplication in $V$.
[/step]
[step:Compare the two curvature matrices in a unitary frame]
Let $U\subset M$ be an [open set](/page/Open%20Set) over which $V$ admits a smooth local $h$-unitary frame $(e_1,\dots,e_r)$. Let
\begin{align*}
\theta=(\theta_{ab})_{1\le a,b\le r}
\end{align*}
be the matrix of complex-valued $1$-forms on $U$ defined by
\begin{align*}
\nabla e_b=\sum_{a=1}^r e_a\,\theta_{ab}.
\end{align*}
Let
\begin{align*}
\Omega=(\Omega_{ab})_{1\le a,b\le r}
\end{align*}
be the curvature matrix of $\nabla$ in this frame, defined by
\begin{align*}
\nabla^2 e_b=\sum_{a=1}^r e_a\,\Omega_{ab}.
\end{align*}
Equivalently,
\begin{align*}
\Omega=d\theta+\theta\wedge\theta.
\end{align*}
Since the frame is unitary and $\nabla$ is compatible with $h$, the connection matrix is skew-Hermitian:
\begin{align*}
\overline{\theta_{ba}}=-\theta_{ab}.
\end{align*}
Taking exterior derivatives and using the matrix wedge product gives the same skew-Hermitian relation for curvature:
\begin{align*}
\overline{\Omega_{ba}}=-\Omega_{ab}.
\end{align*}
Equivalently,
\begin{align*}
\overline\Omega=-\Omega^\top.
\end{align*}
The local frame $(\overline e_1,\dots,\overline e_r)$ of $\overline V$ has connection matrix $\overline\theta$ for $\overline\nabla$, and therefore curvature matrix
\begin{align*}
\overline\Omega=d\overline\theta+\overline\theta\wedge\overline\theta.
\end{align*}
Thus, in a unitary frame, the curvature matrix of the conjugate connection is
\begin{align*}
\Omega_{\overline\nabla}=\overline\Omega=-\Omega^\top.
\end{align*}
[/step]
[step:Compute the determinant Chern form after conjugation]
Let $I_r$ denote the $r\times r$ identity matrix. Define the total Chern-Weil form of $\nabla$ on $U$ by
\begin{align*}
c(V,\nabla):=\det\left(I_r+\frac{i}{2\pi}\Omega\right).
\end{align*}
Because the entries of $\Omega$ are $2$-forms, they commute in the graded-commutative algebra of differential forms, so the ordinary determinant polynomial applies without sign ambiguity.
For the conjugate connection, the preceding step gives
\begin{align*}
c(\overline V,\overline\nabla)=\det\left(I_r+\frac{i}{2\pi}\overline\Omega\right)=\det\left(I_r-\frac{i}{2\pi}\Omega^\top\right).
\end{align*}
Since determinants are invariant under transpose over a commutative algebra,
\begin{align*}
\det\left(I_r-\frac{i}{2\pi}\Omega^\top\right)=\det\left(I_r-\frac{i}{2\pi}\Omega\right).
\end{align*}
Write
\begin{align*}
\det\left(I_r+\frac{i}{2\pi}\Omega\right)=\sum_{j=0}^r c_j(V,\nabla)
\end{align*}
where $c_j(V,\nabla)\in\Omega^{2j}(U;\mathbb C)$ is the homogeneous degree $j$ part in the matrix entries of $\frac{i}{2\pi}\Omega$. Replacing $\frac{i}{2\pi}\Omega$ by $-\frac{i}{2\pi}\Omega$ multiplies the homogeneous degree $j$ part by $(-1)^j$. Hence
\begin{align*}
c_j(\overline V,\overline\nabla)=(-1)^j c_j(V,\nabla)
\end{align*}
on $U$ for every $j\in\{0,\dots,r\}$.
[guided]
We now compare the actual Chern-Weil forms, not just the curvature matrices. Let $I_r$ be the $r\times r$ identity matrix. The total Chern-Weil form of the connection $\nabla$ in the chosen unitary frame is
\begin{align*}
c(V,\nabla):=\det\left(I_r+\frac{i}{2\pi}\Omega\right).
\end{align*}
This determinant is legitimate as an ordinary determinant polynomial because each entry of $\Omega$ is a $2$-form. Differential forms of even degree commute with each other in the graded-commutative algebra $\Omega^*(U;\mathbb C)$, so no extra signs appear when expanding the determinant.
For the conjugate connection $\overline\nabla$, the curvature matrix computed in the previous step is $\overline\Omega=-\Omega^\top$. Therefore
\begin{align*}
c(\overline V,\overline\nabla)=\det\left(I_r+\frac{i}{2\pi}\overline\Omega\right)=\det\left(I_r-\frac{i}{2\pi}\Omega^\top\right).
\end{align*}
The determinant of a matrix equals the determinant of its transpose over a commutative algebra, and the relevant algebra here is commutative for these even-degree entries. Hence
\begin{align*}
\det\left(I_r-\frac{i}{2\pi}\Omega^\top\right)=\det\left(I_r-\frac{i}{2\pi}\Omega\right).
\end{align*}
Now expand the determinant by homogeneous degree. Define $c_j(V,\nabla)\in\Omega^{2j}(U;\mathbb C)$ by
\begin{align*}
\det\left(I_r+\frac{i}{2\pi}\Omega\right)=\sum_{j=0}^r c_j(V,\nabla).
\end{align*}
The term $c_j(V,\nabla)$ is homogeneous of degree $j$ in the matrix entries of $\frac{i}{2\pi}\Omega$. When every entry of $\frac{i}{2\pi}\Omega$ is replaced by its negative, every monomial of degree $j$ is multiplied by $(-1)^j$. Thus the degree $j$ term in
\begin{align*}
\det\left(I_r-\frac{i}{2\pi}\Omega\right)
\end{align*}
is exactly $(-1)^j c_j(V,\nabla)$. Therefore
\begin{align*}
c_j(\overline V,\overline\nabla)=(-1)^j c_j(V,\nabla)
\end{align*}
on $U$ for every $j\in\{0,\dots,r\}$.
[/guided]
[/step]
[step:Pass from local Chern-Weil forms to de Rham Chern classes]
The preceding identity is independent of the chosen unitary frame, because both sides are locally defined components of the globally defined determinant Chern-Weil forms. Hence it holds globally:
\begin{align*}
c_j(\overline V,\overline\nabla)=(-1)^j c_j(V,\nabla)
\end{align*}
in $\Omega^{2j}(M;\mathbb C)$ for every $j\in\{0,\dots,r\}$.
By [citetheorem:9769], the de Rham Chern class $c_j(V)$ is represented by the closed Chern-Weil form $c_j(V,\nabla)$, and $c_j(\overline V)$ is represented by $c_j(\overline V,\overline\nabla)$. Taking cohomology classes gives
\begin{align*}
c_j(\overline V)=(-1)^j c_j(V)
\end{align*}
in $H_{\mathrm{dR}}^{2j}(M;\mathbb C)$ for every $j\in\{0,\dots,r\}$.
For $j>r$, both $c_j(V)$ and $c_j(\overline V)$ vanish by the determinant definition of Chern classes for a rank-$r$ bundle. For $j=0$, both sides equal $1\in H_{\mathrm{dR}}^0(M;\mathbb C)$, and the displayed formula gives $1=(-1)^0 1$. Therefore
\begin{align*}
c_k(\overline V)=(-1)^k c_k(V)
\end{align*}
for every integer $k\ge 0$.
[/step]
