[proofplan]
We first identify the [characteristic polynomial](/page/Characteristic%20Polynomial) of the lattice of flats with the usual subset expansion for the matroid:
\begin{align*}
\chi_{L(M)}(t)=\sum_{A\subset E}(-1)^{|A|}t^{r-\rho(A)}.
\end{align*}
Then we cancel all subsets containing a broken circuit by a rank-preserving sign-reversing involution. The only surviving subsets are the no-broken-circuit subsets, and these are independent, so their rank equals their cardinality. Comparing coefficients of $t^{r-i}$ gives the stated formula for $w_i$.
[/proofplan]
[step:Derive the subset expansion from the flat lattice Möbius function]
The closure map declared in the theorem statement is
\begin{align*}
\operatorname{cl}_M:2^E&\to L(M)
\end{align*}
\begin{align*}
A&\mapsto \operatorname{cl}_M(A).
\end{align*}
Since $M$ is loopless, $\operatorname{cl}_M(\varnothing)=\varnothing$, so $\varnothing$ is the minimum element of $L(M)$.
Define the integer-valued map
\begin{align*}
S:L(M)&\to \mathbb Z
\end{align*}
\begin{align*}
F&\mapsto \sum_{\substack{A\subset E, \operatorname{cl}_M(A)=F}}(-1)^{|A|}.
\end{align*}
If $F\in L(M)$, then the flats $G\subset F$ are exactly the possible closures of subsets $A\subset F$. Hence
\begin{align*}
\sum_{\substack{G\in L(M), G\subset F}}S(G)=\sum_{A\subset F}(-1)^{|A|}.
\end{align*}
The right-hand side is $(1-1)^{|F|}$. Therefore it is $1$ when $F=\varnothing$ and $0$ when $F\ne\varnothing$.
By [citetheorem:8094] on the finite poset $L(M)$, equivalently by the defining recursion for the Möbius function, this gives
\begin{align*}
S(F)=\mu_{L(M)}(\varnothing,F)
\end{align*}
for every flat $F\in L(M)$. Since $\rho(A)=\rho(\operatorname{cl}_M(A))$, we obtain
\begin{align*}
\chi_{L(M)}(t)=\sum_{F\in L(M)}\mu_{L(M)}(\varnothing,F)t^{r-\rho(F)}=\sum_{A\subset E}(-1)^{|A|}t^{r-\rho(A)}.
\end{align*}
[guided]
The characteristic polynomial of $L(M)$ is defined by summing the Möbius function over flats:
\begin{align*}
\chi_{L(M)}(t)=\sum_{F\in L(M)}\mu_{L(M)}(\varnothing,F)t^{r-\rho(F)}.
\end{align*}
To rewrite this as a sum over all subsets of $E$, we group subsets according to their closure. Define the map
\begin{align*}
S:L(M)&\to \mathbb Z
\end{align*}
\begin{align*}
F&\mapsto \sum_{\substack{A\subset E, \operatorname{cl}_M(A)=F}}(-1)^{|A|}.
\end{align*}
We want to prove that $S(F)$ is exactly $\mu_{L(M)}(\varnothing,F)$.
Fix a flat $F$. A subset $A\subset E$ has closure contained in $F$ if and only if $A\subset F$, because $F$ is closed. Therefore, summing $S(G)$ over all flats $G\subset F$ counts every subset $A\subset F$ exactly once, namely under the flat $G=\operatorname{cl}_M(A)$. Thus
\begin{align*}
\sum_{\substack{G\in L(M), G\subset F}}S(G)=\sum_{A\subset F}(-1)^{|A|}.
\end{align*}
The last sum is the binomial expansion of $(1-1)^{|F|}$. It equals $1$ when $F=\varnothing$ and equals $0$ when $F\ne\varnothing$.
Because $M$ is loopless, $\operatorname{cl}_M(\varnothing)=\varnothing$, so $\varnothing$ is the minimum flat. The defining recursion for the Möbius function on the finite poset $L(M)$, equivalently [citetheorem:8094] applied to the interval from $\varnothing$ to $F$, says that the unique function $F\mapsto \mu_{L(M)}(\varnothing,F)$ satisfies
\begin{align*}
\sum_{\substack{G\in L(M), G\subset F}}\mu_{L(M)}(\varnothing,G)=1
\end{align*}
when $F=\varnothing$, and satisfies
\begin{align*}
\sum_{\substack{G\in L(M), G\subset F}}\mu_{L(M)}(\varnothing,G)=0
\end{align*}
when $F\ne\varnothing$. The function $S$ satisfies the same equations, hence $S(F)=\mu_{L(M)}(\varnothing,F)$ for every flat $F$.
Finally, rank is unchanged by closure, so $\rho(A)=\rho(\operatorname{cl}_M(A))$. Substituting the identity for $S(F)$ into the characteristic polynomial gives
\begin{align*}
\chi_{L(M)}(t)=\sum_{F\in L(M)}\mu_{L(M)}(\varnothing,F)t^{r-\rho(F)}=\sum_{A\subset E}(-1)^{|A|}t^{r-\rho(A)}.
\end{align*}
[/guided]
[/step]
[step:Define the broken-circuit cancellation domain]
Let $\mathcal B$ be the set of all subsets $A\subset E$ that contain a broken circuit. Thus $A\in\mathcal B$ if and only if there exists a circuit $C$ of $M$ such that
\begin{align*}
C\setminus\{\min_{<}C\}\subset A.
\end{align*}
Define the map
\begin{align*}
e:\mathcal B&\to E
\end{align*}
by letting $e(A)$ be the smallest element $e\in E$ for which there exists a circuit $C$ of $M$ satisfying
\begin{align*}
\min_{<}C=e
\end{align*}
and
\begin{align*}
C\setminus\{e\}\subset A.
\end{align*}
Such an element exists by the definition of $\mathcal B$, and it is unique because $E$ is finite and linearly ordered. Choose one circuit with this property and denote it by $C_A$.
Define a map
\begin{align*}
\Phi:\mathcal B&\to 2^E
\end{align*}
by
\begin{align*}
\Phi(A)=A\triangle\{e(A)\},
\end{align*}
where $\triangle$ denotes symmetric difference. Equivalently, $\Phi(A)$ is obtained from $A$ by toggling the single element $e(A)$.
[/step]
[step:Show that toggling preserves the chosen broken-circuit witness]
We claim that $\Phi(A)\in\mathcal B$ and $e(\Phi(A))=e(A)$ for every $A\in\mathcal B$.
Let $e=e(A)$ and $C=C_A$. Since $e=\min_{<}C$, the element $e$ is not contained in $C\setminus\{e\}$. Hence
\begin{align*}
C\setminus\{e\}\subset A\triangle\{e\}=\Phi(A).
\end{align*}
Thus $\Phi(A)\in\mathcal B$, and $e$ is a candidate for $e(\Phi(A))$.
It remains to prove that no smaller element is a candidate for $\Phi(A)$. Suppose, for contradiction, that there exist an element $d<e$ and a circuit $D$ such that $\min_{<}D=d$ and
\begin{align*}
D\setminus\{d\}\subset \Phi(A).
\end{align*}
If $e\notin D\setminus\{d\}$, then $D\setminus\{d\}\subset A$, contradicting the minimality of $e=e(A)$.
Thus $e\in D$. Since $d<e=\min_{<}C$, we have $d\notin C$. The strong circuit elimination axiom, applied to the circuits $C$ and $D$, the common element $e\in C\cap D$, and the element $d\in D\setminus C$, gives a circuit $H$ satisfying
\begin{align*}
d\in H\subset (C\cup D)\setminus\{e\}.
\end{align*}
Every element of $C\setminus\{e\}$ lies in $A$, and every element of $D\setminus\{d,e\}$ lies in $A$. Therefore
\begin{align*}
H\setminus\{d\}\subset A.
\end{align*}
Moreover $d=\min_{<}H$, because all elements of $C\setminus\{e\}$ are greater than $e$ and all elements of $D\setminus\{d\}$ are greater than $d$. Hence $d$ is a candidate for $e(A)$, contradicting $d<e(A)$.
Therefore no candidate smaller than $e$ exists for $\Phi(A)$, and so
\begin{align*}
e(\Phi(A))=e(A).
\end{align*}
[guided]
This is the delicate point of the proof. We define the involution by toggling the smallest possible first element of a broken-circuit witness, and we must prove that this smallest element does not change after the toggle.
Fix $A\in\mathcal B$. Write $e=e(A)$, and choose a circuit $C=C_A$ such that $\min_{<}C=e$ and
\begin{align*}
C\setminus\{e\}\subset A.
\end{align*}
The set $C\setminus\{e\}$ does not contain $e$, so toggling $e$ does not affect this containment. Hence
\begin{align*}
C\setminus\{e\}\subset A\triangle\{e\}=\Phi(A).
\end{align*}
This proves that $\Phi(A)$ still contains a broken circuit, and that $e$ is still a possible choice for the first element of a witnessing circuit.
Now suppose that after toggling, a smaller element $d<e$ became possible. Then there is a circuit $D$ with $\min_{<}D=d$ and
\begin{align*}
D\setminus\{d\}\subset \Phi(A).
\end{align*}
If $e$ is not in $D\setminus\{d\}$, then toggling $e$ did not affect the set $D\setminus\{d\}$, so we would have
\begin{align*}
D\setminus\{d\}\subset A.
\end{align*}
This would make $d$ a valid candidate for $e(A)$, contradicting the minimality of $e(A)=e$.
Therefore the only possible way a new smaller candidate could appear is if $e\in D$. We then use circuit elimination. The circuits $C$ and $D$ both contain $e$, and $d\in D\setminus C$ because $d<e=\min_{<}C$. The strong circuit elimination axiom gives a circuit $H$ such that
\begin{align*}
d\in H\subset (C\cup D)\setminus\{e\}.
\end{align*}
The elements of $C\setminus\{e\}$ already lie in $A$, and the elements of $D\setminus\{d,e\}$ also lie in $A$ because they are not the toggled element. Hence all elements of $H$ except possibly $d$ lie in $A$:
\begin{align*}
H\setminus\{d\}\subset A.
\end{align*}
Since $d$ is the least element of $D$ and every element of $C\setminus\{e\}$ is greater than $e>d$, we also have $\min_{<}H=d$. Thus $d$ was already a valid candidate for $e(A)$ before the toggle, again contradicting $d<e(A)$.
This contradiction proves that no smaller candidate appears after toggling. Since $e$ remains a candidate, we conclude
\begin{align*}
e(\Phi(A))=e(A).
\end{align*}
[/guided]
[/step]
[step:Use the toggle map as a rank-preserving sign-reversing involution]
For $A\in\mathcal B$, let $e=e(A)$ and choose $C=C_A$ as above. Since
\begin{align*}
C\setminus\{e\}\subset A
\end{align*}
and $C$ is a circuit, the element $e$ lies in the closure of $C\setminus\{e\}$, hence also in the closure of $A\setminus\{e\}$ whenever $e\in A$, and in the closure of $A$ whenever $e\notin A$. Therefore toggling $e$ does not change rank:
\begin{align*}
\rho(\Phi(A))=\rho(A).
\end{align*}
The previous step gives $e(\Phi(A))=e(A)$, so toggling once more returns to $A$:
\begin{align*}
\Phi(\Phi(A))=A.
\end{align*}
Also, $A$ and $\Phi(A)$ differ by exactly one element, so
\begin{align*}
(-1)^{|\Phi(A)|}=-(-1)^{|A|}.
\end{align*}
Thus $\Phi$ is a rank-preserving sign-reversing involution on $\mathcal B$. Consequently,
\begin{align*}
\sum_{A\in\mathcal B}(-1)^{|A|}t^{r-\rho(A)}=0.
\end{align*}
[/step]
[step:Identify the surviving subsets and compare coefficients]
Let $\mathcal N$ be the set of no-broken-circuit subsets of $E$. Since every subset of $E$ either lies in $\mathcal B$ or in $\mathcal N$, the subset expansion and the cancellation just proved give
\begin{align*}
\chi_{L(M)}(t)=\sum_{A\in\mathcal N}(-1)^{|A|}t^{r-\rho(A)}.
\end{align*}
Every no-broken-circuit subset is independent. Indeed, if $A\in\mathcal N$ were dependent, then $A$ would contain a circuit $C$, and therefore it would contain the broken circuit $C\setminus\{\min_{<}C\}$, contradicting the definition of $\mathcal N$.
Thus $\rho(A)=|A|$ for every $A\in\mathcal N$, and so
\begin{align*}
\chi_{L(M)}(t)=\sum_{A\in\mathcal N}(-1)^{|A|}t^{r-|A|}.
\end{align*}
Grouping this sum by cardinality gives
\begin{align*}
\chi_{L(M)}(t)=\sum_{i=0}^{r}(-1)^i |\{A\in\mathcal N:|A|=i\}|t^{r-i}.
\end{align*}
Comparing this expression with
\begin{align*}
\chi_{L(M)}(t)=\sum_{i=0}^{r}(-1)^i w_i t^{r-i}
\end{align*}
shows that
\begin{align*}
w_i=|\{A\subset E:A\text{ is no-broken-circuit and }|A|=i\}|
\end{align*}
for every $0\le i\le r$. This is the desired coefficient formula.
[/step]
