[proofplan]
We write the Chern character form as a finite sum of Chern-Weil forms associated to the invariant polynomials $A\mapsto \frac{1}{m!}\operatorname{tr}((A/2\pi i)^m)$. Chern-Weil closedness gives closedness term by term, and the [Chern-Weil homotopy formula](/theorems/9760) gives connection independence for each homogeneous component. For the root formula, we pull back along a splitting map, compute the formula for a [direct sum](/page/Direct%20Sum) of line bundles, and descend the equality using the injectivity part of the splitting principle.
[/proofplan]
[step:Express the Chern character as a sum of invariant polynomial Chern-Weil forms]
Let
\begin{align*}
\operatorname{Ad}:GL(r,\mathbb C)\to GL(\mathfrak{gl}(r,\mathbb C))
\end{align*}
denote the adjoint representation, defined by
\begin{align*}
\operatorname{Ad}_B A:=BAB^{-1}
\end{align*}
for $B\in GL(r,\mathbb C)$ and $A\in\mathfrak{gl}(r,\mathbb C)$.
For each integer $m\ge 0$, define the polynomial
\begin{align*}
P_m:\mathfrak{gl}(r,\mathbb C)&\to \mathbb C
\end{align*}
by
\begin{align*}
P_m(A)=\frac{1}{m!}\operatorname{tr}\left(\left(\frac{A}{2\pi i}\right)^m\right)
\end{align*}
for $A\in\mathfrak{gl}(r,\mathbb C)$, with the convention $P_0(A)=r$. The trace identity $\operatorname{tr}(BAB^{-1})=\operatorname{tr}(A)$ for $B\in GL(r,\mathbb C)$ implies
\begin{align*}
P_m(\operatorname{Ad}_B A)=P_m(A),
\end{align*}
so $P_m$ is an $\operatorname{Ad}$-invariant [homogeneous polynomial](/page/Homogeneous%20Polynomial) of degree $m$ for $m\ge 1$.
Let $n:=\dim M$. Since $F_{\nabla^E}$ has form degree $2$, the component of degree $2m$ vanishes for $2m>n$. Hence
\begin{align*}
\operatorname{ch}(E,\nabla^E)=\sum_{0\le m\le n/2}P_m(F_{\nabla^E})
\end{align*}
as an element of $\Omega^{\mathrm{even}}(M;\mathbb C)$.
[/step]
[step:Apply Chern-Weil closedness to every homogeneous component]
Let $\operatorname{Fr}(E)\to M$ be the principal $GL(r,\mathbb C)$-bundle of complex frames of $E$. The connection $\nabla^E$ corresponds to a principal connection on $\operatorname{Fr}(E)$, and its curvature corresponds to $F_{\nabla^E}$ under the standard representation of $GL(r,\mathbb C)$ on $\mathbb C^r$. Since each $P_m$ is $\operatorname{Ad}$-invariant, [citetheorem:9757] applied to the frame bundle gives
\begin{align*}
d\bigl(P_m(F_{\nabla^E})\bigr)=0
\end{align*}
for every $m\ge 0$. Therefore the finite sum $\operatorname{ch}(E,\nabla^E)$ is closed:
\begin{align*}
d\,\operatorname{ch}(E,\nabla^E)=0.
\end{align*}
[guided]
The point of rewriting the Chern character as a sum of the forms $P_m(F_{\nabla^E})$ is that Chern-Weil theory applies exactly to invariant polynomials evaluated on curvature. For each integer $m\ge 0$, the polynomial
\begin{align*}
P_m(A)=\frac{1}{m!}\operatorname{tr}\left(\left(\frac{A}{2\pi i}\right)^m\right)
\end{align*}
is invariant under conjugation because the trace is invariant under conjugation:
\begin{align*}
\operatorname{tr}\left(\left(\frac{BAB^{-1}}{2\pi i}\right)^m\right)
=
\operatorname{tr}\left(B\left(\frac{A}{2\pi i}\right)^mB^{-1}\right)
=
\operatorname{tr}\left(\left(\frac{A}{2\pi i}\right)^m\right).
\end{align*}
The vector bundle connection $\nabla^E$ is equivalently a principal connection on the frame bundle $\operatorname{Fr}(E)\to M$ with structure group $GL(r,\mathbb C)$. Under the standard representation, the principal curvature form is represented on $E$ by the endomorphism-valued curvature form $F_{\nabla^E}$. Thus the hypotheses of [citetheorem:9757] are satisfied: the structure group is $GL(r,\mathbb C)$, the connection is the frame-bundle connection induced by $\nabla^E$, and the polynomial $P_m$ is $\operatorname{Ad}$-invariant. Consequently
\begin{align*}
d\bigl(P_m(F_{\nabla^E})\bigr)=0
\end{align*}
for each $m$.
Because $F_{\nabla^E}$ has degree $2$, only finitely many terms contribute on the finite-dimensional manifold $M$. The [exterior derivative](/theorems/1525) is linear, so summing the closed homogeneous components gives
\begin{align*}
d\,\operatorname{ch}(E,\nabla^E)
=
d\left(\sum_{0\le m\le n/2}P_m(F_{\nabla^E})\right)
=
\sum_{0\le m\le n/2}d\bigl(P_m(F_{\nabla^E})\bigr)
=
0.
\end{align*}
Thus the Chern character form is closed.
[/guided]
[/step]
[step:Use transgression to prove independence of the connection]
Let $\nabla^0$ and $\nabla^1$ be two smooth complex-linear connections on $E$. Define
\begin{align*}
A:=\nabla^1-\nabla^0\in\Omega^1(M;\operatorname{End}(E))
\end{align*}
and, for $t\in[0,1]$, define the affine path of connections
\begin{align*}
\nabla^t:=\nabla^0+tA.
\end{align*}
For each $t\in[0,1]$, the operator $\nabla^t=\nabla^0+tA$ is a smooth complex-linear connection on the same vector bundle $E\to M$, because the difference of two complex-linear connections is an element of $\Omega^1(M;\operatorname{End}(E))$. For each $m\ge 1$, the polynomial $P_m$ has already been shown to be $\operatorname{Ad}$-invariant, so the hypotheses of [citetheorem:9760] are satisfied for the affine path $(\nabla^t)_{t\in[0,1]}$. Thus [citetheorem:9760] gives a form
\begin{align*}
T_m\in\Omega^{2m-1}(M;\mathbb C)
\end{align*}
such that
\begin{align*}
P_m(F_{\nabla^1})-P_m(F_{\nabla^0})=dT_m.
\end{align*}
For $m=0$, the term $P_0(F_{\nabla^t})=r$ is constant, so its difference is zero. Summing over the finitely many nonzero form degrees gives
\begin{align*}
\operatorname{ch}(E,\nabla^1)-\operatorname{ch}(E,\nabla^0)
=
d\left(\sum_{1\le m\le n/2}T_m\right).
\end{align*}
Therefore $\operatorname{ch}(E,\nabla^1)$ and $\operatorname{ch}(E,\nabla^0)$ define the same class in $H_{\mathrm{dR}}^{\mathrm{even}}(M;\mathbb C)$, so $\operatorname{ch}(E)$ is well-defined.
[/step]
[step:Compute the Chern character after splitting into line bundles]
We now use the splitting principle for complex vector bundles, stated here as the standard prerequisite that there exists a smooth manifold $Y$ and a smooth map
\begin{align*}
s:Y\to M
\end{align*}
such that the induced pullback homomorphism
\begin{align*}
s^*:H_{\mathrm{dR}}^*(M;\mathbb C)\to H_{\mathrm{dR}}^*(Y;\mathbb C)
\end{align*}
is injective and such that the pulled-back complex vector bundle $s^*E\to Y$ admits a complex vector bundle isomorphism
\begin{align*}
s^*E\cong L_1\oplus\cdots\oplus L_r
\end{align*}
for smooth complex line bundles $L_1,\dots,L_r\to Y$. For each $j\in\{1,\dots,r\}$, let $c_1(L_j)\in H_{\mathrm{dR}}^2(Y;\mathbb C)$ denote the first Chern class of the complex line bundle $L_j\to Y$, and define
\begin{align*}
y_j:=c_1(L_j)\in H_{\mathrm{dR}}^2(Y;\mathbb C).
\end{align*}
By [naturality of Chern classes](/theorems/9771) under pullback, as in [citetheorem:9771], and by the [Whitney product formula for Chern classes](/theorems/7052) of the split bundle,
\begin{align*}
s^*c(E)=c(s^*E)=\prod_{j=1}^r(1+y_j).
\end{align*}
Equivalently, the elementary symmetric polynomial $e_k(y_1,\dots,y_r)$ equals $s^*c_k(E)$ for every $k\in\{0,\dots,r\}$.
Choose smooth complex-linear connections $\nabla^{L_j}$ on $L_j$, and equip $s^*E\cong\bigoplus_{j=1}^rL_j$ with the direct-sum connection
\begin{align*}
\nabla^\oplus:=\nabla^{L_1}\oplus\cdots\oplus\nabla^{L_r}.
\end{align*}
Its curvature is block diagonal:
\begin{align*}
F_{\nabla^\oplus}=F_{\nabla^{L_1}}\oplus\cdots\oplus F_{\nabla^{L_r}}.
\end{align*}
Since the trace of a block diagonal endomorphism is the sum of the traces of its blocks,
\begin{align*}
\operatorname{ch}(s^*E,\nabla^\oplus)
=
\sum_{j=1}^r \operatorname{ch}(L_j,\nabla^{L_j}).
\end{align*}
For a line bundle $L_j$, the curvature $F_{\nabla^{L_j}}\in\Omega^2(Y;\operatorname{End}(L_j))$ acts by a scalar-valued $2$-form, so
\begin{align*}
\operatorname{ch}(L_j,\nabla^{L_j})
=
\exp\left(\frac{F_{\nabla^{L_j}}}{2\pi i}\right).
\end{align*}
By the [Chern-Weil construction of Chern classes](/theorems/9769), as in [citetheorem:9769], the normalization of the first Chern form identifies
\begin{align*}
c_1(L_j)=\left[\frac{F_{\nabla^{L_j}}}{2\pi i}\right]\in H_{\mathrm{dR}}^2(Y;\mathbb C).
\end{align*}
Therefore, passing to cohomology and using the multiplicative definition of the exponential of an even cohomology class,
\begin{align*}
\operatorname{ch}(L_j)
=
\left[\exp\left(\frac{F_{\nabla^{L_j}}}{2\pi i}\right)\right]
=
\exp(c_1(L_j))
=
e^{y_j}.
\end{align*}
Thus
\begin{align*}
\operatorname{ch}(s^*E)=\sum_{j=1}^r e^{y_j}.
\end{align*}
[/step]
[step:Descend the split identity to the original bundle]
By [naturality of Chern-Weil forms](/theorems/9761) under pullback, as in [citetheorem:9761], the pulled-back connection $s^*\nabla^E$ satisfies
\begin{align*}
\operatorname{ch}(s^*E,s^*\nabla^E)=s^*\operatorname{ch}(E,\nabla^E).
\end{align*}
Passing to cohomology gives
\begin{align*}
\operatorname{ch}(s^*E)=s^*\operatorname{ch}(E).
\end{align*}
Let $\Phi$ denote the finite symmetric polynomial in $r$ formal variables obtained from the degree-at-most-$n$ truncation of $\sum_{j=1}^r e^{z_j}$. By the [fundamental theorem of symmetric polynomials](/theorems/5179), $\Phi$ can be written uniquely as a polynomial $\Psi$ in the elementary symmetric polynomials. Therefore the definition of the formal-root expression gives
\begin{align*}
\sum_{j=1}^r e^{x_j}=\Psi(c_1(E),\dots,c_r(E)).
\end{align*}
Since $e_k(y_1,\dots,y_r)=s^*c_k(E)$ for every $k$, the split computation gives
\begin{align*}
s^*\operatorname{ch}(E)
=
\operatorname{ch}(s^*E)
=
\sum_{j=1}^r e^{y_j}
=
\Psi(s^*c_1(E),\dots,s^*c_r(E))
=
s^*\left(\sum_{j=1}^r e^{x_j}\right).
\end{align*}
Since the splitting principle gives that $s^*$ is injective on cohomology, we conclude
\begin{align*}
\operatorname{ch}(E)=\sum_{j=1}^r e^{x_j}
\end{align*}
in $H_{\mathrm{dR}}^{\mathrm{even}}(M;\mathbb C)$. This proves the closedness, connection independence, and formal Chern-root formula for the Chern character.
[/step]
