[proofplan]
We prove the alternating cancellation on an arbitrary nontrivial interval $[F,G]$ of the face lattice. The main point is that every such interval is itself the face lattice of a lower-dimensional convex polytope, namely the link of $F$ inside $G$, with ranks shifted by $\rho(F)$. Applying the classical Euler-Poincare relation for convex polytopes to that link gives the desired cancellation. Since the interval was arbitrary, the face lattice is Eulerian.
[/proofplan]
[step:Fix the rank convention and reduce to one interval]
Let $F,G\in L(Q)$ satisfy $F<G$, meaning $F\subset G$ as faces of $Q$. Thus the interval is
\begin{align*}
[F,G]=\{H\in L(Q):F\subseteq H\subseteq G\}.
\end{align*}
We use the rank convention from the statement:
\begin{align*}
\rho(\varnothing)=0
\end{align*}
and, for every nonempty face $H$,
\begin{align*}
\rho(H)=\dim H+1.
\end{align*}
It is enough to prove
\begin{align*}
\sum_{H\in [F,G]}(-1)^{\rho(H)-\rho(F)}=0
\end{align*}
for this fixed interval, because the Eulerian condition requires exactly this cancellation on every nontrivial interval.
[/step]
[step:Construct a face figure whose face lattice is the interval]
Since $F<G$ in $L(Q)$, the element $F$ is a face of the polytope $G$. We use the standard face-figure theorem for convex polytopes, in its interval form: if $F$ is a face of a convex polytope $G$, then there is a convex polytope $K$, called a face figure or link of $F$ in $G$, of dimension
\begin{align*}
\rho(G)-\rho(F)-1
\end{align*}
whose face lattice is order-isomorphic to the interval $[F,G]$, with the empty face of $K$ corresponding to $F$ and the full face $K$ corresponding to $G$.
More concretely, when $F=\varnothing$, take $K=G$ and let $\Phi:L(K)\to [F,G]$ be the identity map. When $F\neq\varnothing$, choose a point $x_0$ in the relative interior of $F$, choose an affine subspace through $x_0$ transverse to the affine hull of $F$ inside the affine hull of $G$, and choose an exposing affine functional for $F$ in $G$. For sufficiently small positive level, the transverse section is a convex polytope $K$; its faces are exactly the nonempty intersections with faces $H$ satisfying $F<H\le G$, together with the empty face corresponding to $F$. This gives an order-isomorphism
\begin{align*}
\Phi:L(K)\to [F,G].
\end{align*}
Let $\varnothing_K$ denote the empty face of $K$. The isomorphism satisfies
\begin{align*}
\Phi(\varnothing_K)=F
\end{align*}
and
\begin{align*}
\Phi(K)=G.
\end{align*}
Let $\rho_K:L(K)\to \mathbb{N}\cup\{0\}$ denote the face-lattice rank function on $L(K)$, normalized by $\rho_K(\varnothing_K)=0$ and $\rho_K(A)=\dim A+1$ for every nonempty face $A$ of $K$. The face-figure construction decreases all ranks by $\rho(F)$, so for every face $A\in L(K)$,
\begin{align*}
\rho_K(A)=\rho(\Phi(A))-\rho(F).
\end{align*}
[guided]
The purpose of this step is to replace an arbitrary interval by an entire face lattice, because the Euler-Poincare formula is naturally stated for whole polytopes. We first check the hypothesis needed for the face-figure construction: since $F<G$ in the face lattice $L(Q)$, the face $F$ is contained in the face $G$, and hence $F$ is a face of the convex polytope $G$.
The standard face-figure theorem for convex polytopes, in its interval form, says that if $F$ is a face of a convex polytope $G$, then there is a convex polytope $K$, called the face figure or link of $F$ in $G$, whose face lattice is order-isomorphic to the interval above $F$ and below $G$. Its dimension is
\begin{align*}
\rho(G)-\rho(F)-1.
\end{align*}
This dimension formula matches the rank convention: the top element of $L(K)$ has rank $\dim K+1$, and the interval $[F,G]$ has relative top rank $\rho(G)-\rho(F)$.
When $F=\varnothing$, no transverse construction is needed: the interval $[\varnothing,G]$ is already the full face lattice of $G$, so we take $K=G$ and use the identity correspondence. When $F\neq\varnothing$, one may construct $K$ as a small transverse section near a point $x_0$ in the relative interior of $F$. Choose an affine subspace through $x_0$ transverse to the affine hull of $F$ inside the affine hull of $G$, and choose an affine functional exposing $F$ in $G$. Intersecting $G$ with the transverse affine subspace and with a sufficiently small positive level of the exposing functional produces a convex polytope $K$. The faces of this section are exactly the intersections coming from faces $H$ with $F<H\le G$; the missing lower endpoint $F$ is represented by the empty face of $K$.
Thus there is an order-isomorphism
\begin{align*}
\Phi:L(K)\to [F,G]
\end{align*}
with
\begin{align*}
\Phi(\varnothing_K)=F
\end{align*}
and
\begin{align*}
\Phi(K)=G,
\end{align*}
where $\varnothing_K$ denotes the empty face of $K$. Let $\rho_K:L(K)\to \mathbb{N}\cup\{0\}$ be the rank function on $L(K)$, normalized by $\rho_K(\varnothing_K)=0$ and $\rho_K(A)=\dim A+1$ for nonempty faces $A$ of $K$. Since the face figure records exactly how far a face lies above $F$, the rank in $L(K)$ is the relative rank in the interval:
\begin{align*}
\rho_K(A)=\rho(\Phi(A))-\rho(F)
\end{align*}
for every face $A\in L(K)$. This identity is the sign bookkeeping needed to transfer the alternating sum from $[F,G]$ to $L(K)$.
[/guided]
[/step]
[step:Apply Euler-Poincare cancellation to the link polytope]
We use the independently established classical Euler-Poincare formula for convex polytopes in the following precise form: if $P$ is a convex polytope and $\rho_P$ is the rank function on its face lattice, normalized by $\rho_P(\varnothing_P)=0$ and $\rho_P(B)=\dim B+1$ for nonempty faces $B$, then
\begin{align*}
\sum_{B\in L(P)}(-1)^{\rho_P(B)}=0.
\end{align*}
The hypotheses hold for $P=K$, because the preceding step constructed $K$ as a convex polytope and defined $\rho_K$ with this normalization. Therefore
\begin{align*}
\sum_{A\in L(K)}(-1)^{\rho_K(A)}=0.
\end{align*}
Using the bijection $\Phi:L(K)\to [F,G]$ and the rank identity from the previous step, we obtain
\begin{align*}
\sum_{H\in [F,G]}(-1)^{\rho(H)-\rho(F)}
=
\sum_{A\in L(K)}(-1)^{\rho(\Phi(A))-\rho(F)}.
\end{align*}
Using $\rho_K(A)=\rho(\Phi(A))-\rho(F)$ for every $A\in L(K)$, this becomes
\begin{align*}
\sum_{H\in [F,G]}(-1)^{\rho(H)-\rho(F)}
=
\sum_{A\in L(K)}(-1)^{\rho_K(A)}.
\end{align*}
The Euler-Poincare formula for $K$ gives
\begin{align*}
\sum_{H\in [F,G]}(-1)^{\rho(H)-\rho(F)}=0.
\end{align*}
This includes the rank-one case: then $K$ is a point, and the sum over its two faces is
\begin{align*}
(-1)^0+(-1)^1=0.
\end{align*}
[guided]
We now use the Euler-Poincare formula on the auxiliary polytope $K$. The exact external result needed is the independently established classical Euler-Poincare formula for convex polytopes: if $P$ is a convex polytope and $\rho_P$ is the face-lattice rank function satisfying $\rho_P(\varnothing_P)=0$ and $\rho_P(B)=\dim B+1$ for every nonempty face $B$, then
\begin{align*}
\sum_{B\in L(P)}(-1)^{\rho_P(B)}=0.
\end{align*}
We verify its hypotheses for $P=K$. The previous step constructed $K$ as a convex polytope, and $\rho_K$ was defined with exactly the required rank normalization. Hence the formula applies and gives
\begin{align*}
\sum_{A\in L(K)}(-1)^{\rho_K(A)}=0.
\end{align*}
It remains to translate this equality back to the original interval. The map $\Phi:L(K)\to [F,G]$ is a bijection, so summing over faces $H$ in $[F,G]$ is the same as summing over their unique preimages $A$ in $L(K)$ with $H=\Phi(A)$. Therefore
\begin{align*}
\sum_{H\in [F,G]}(-1)^{\rho(H)-\rho(F)}
=
\sum_{A\in L(K)}(-1)^{\rho(\Phi(A))-\rho(F)}.
\end{align*}
The rank identity from the face-figure construction says that
\begin{align*}
\rho(\Phi(A))-\rho(F)=\rho_K(A)
\end{align*}
for every $A\in L(K)$. Substituting this into the previous sum gives
\begin{align*}
\sum_{H\in [F,G]}(-1)^{\rho(H)-\rho(F)}
=
\sum_{A\in L(K)}(-1)^{\rho_K(A)}.
\end{align*}
The Euler-Poincare formula for $K$ gives
\begin{align*}
\sum_{H\in [F,G]}(-1)^{\rho(H)-\rho(F)}=0.
\end{align*}
This proves the required cancellation for the fixed interval $[F,G]$. In the rank-one case, the auxiliary polytope $K$ is a point, so its face lattice has only $\varnothing_K$ and $K$, and the cancellation is
\begin{align*}
(-1)^0+(-1)^1=0.
\end{align*}
[/guided]
[/step]
[step:Conclude that the face lattice is Eulerian]
The interval $[F,G]$ was an arbitrary interval of $L(Q)$ with $F<G$, and we have proved
\begin{align*}
\sum_{H\in [F,G]}(-1)^{\rho(H)-\rho(F)}=0.
\end{align*}
Therefore every nontrivial interval in $L(Q)$ has equal alternating rank contribution in even and odd relative ranks. Hence $L(Q)$ is Eulerian.
[/step]
