[proofplan] We label each cover relation in the Boolean lattice by the unique element added along that cover. On an interval $[S,T]$, maximal chains are exactly orderings of the finite set $T \setminus S$, and the label word records that ordering. The increasing ordering of $T \setminus S$ is the unique rising label word and is lexicographically first. Finally, we use the induced lexicographic order on maximal chains to shell the order complex of the proper part directly, by the standard adjacent-descent swap argument. [/proofplan]

[step:Define the Boolean edge labelling] Let $[n] := \{1,\dots,n\}$, and let $B_n := \{S : S \subset [n]\}$ be ordered by inclusion. A cover relation in $B_n$ has the form $S \lessdot S \cup \{i\}$, where $S \subset [n]$ and $i \in [n] \setminus S$. Define the edge labelling \begin{align*} \lambda: \{(A,B) \in B_n \times B_n : A \lessdot B\} \to [n], \qquad \lambda(S, S \cup \{i\}) := i. \end{align*} This is well-defined because $S \lessdot U$ in $B_n$ holds exactly when $U = S \cup \{i\}$ for a unique element $i \in [n] \setminus S$. [/step]

[step:Identify maximal chains in each interval with permutations of the added elements]Fix an interval $[S,T] := \{U \in B_n : S \subset U \subset T\}$. Let \begin{align*} T \setminus S = \{a_1,\dots,a_k\}, \end{align*} where $k := |T \setminus S|$. A maximal chain in $[S,T]$ has the form \begin{align*} S = S_0 \lessdot S_1 \lessdot \cdots \lessdot S_k = T. \end{align*} For each $j \in \{1,\dots,k\}$, there is a unique element $i_j \in T \setminus S_{j-1}$ such that $S_j = S_{j-1} \cup \{i_j\}$. Therefore the chain determines the word $(i_1,\dots,i_k)$, which is a permutation of $T \setminus S$. Conversely, every permutation $(i_1,\dots,i_k)$ of $T \setminus S$ determines the maximal chain \begin{align*} S_0 &= S, \qquad S_j = S \cup \{i_1,\dots,i_j\} \quad \text{for } 1 \leq j \leq k. \end{align*} The label word of this chain is exactly $(i_1,\dots,i_k)$, because the edge $S_{j-1} \lessdot S_j$ adds precisely $i_j$.[/step]

[guided]We need to understand the interval $[S,T]$ in completely concrete terms. The elements that must be added in order to move from $S$ to $T$ are precisely the elements of $T \setminus S$. Write \begin{align*} T \setminus S = \{a_1,\dots,a_k\}, \end{align*} where $k = |T \setminus S|$. A maximal chain from $S$ to $T$ must add exactly one new element at each cover step. Thus if \begin{align*} S = S_0 \lessdot S_1 \lessdot \cdots \lessdot S_k = T \end{align*} is a maximal chain, then for every $j \in \{1,\dots,k\}$ there is a unique element $i_j \in T \setminus S_{j-1}$ such that \begin{align*} S_j = S_{j-1} \cup \{i_j\}. \end{align*} No element can be added twice, and every element of $T \setminus S$ must eventually be added, because the final set is $T$. Hence $(i_1,\dots,i_k)$ is a permutation of $T \setminus S$. Conversely, suppose we choose a permutation $(i_1,\dots,i_k)$ of $T \setminus S$. Define sets \begin{align*} S_0 &= S, \qquad S_j = S \cup \{i_1,\dots,i_j\} \quad \text{for } 1 \leq j \leq k. \end{align*} Then each inclusion $S_{j-1} \subset S_j$ adds exactly one element, so $S_{j-1} \lessdot S_j$ in $B_n$. Also $S_k = S \cup (T \setminus S) = T$. Thus this construction gives a maximal chain in $[S,T]$. By the definition of $\lambda$, the label on the edge $S_{j-1} \lessdot S_j$ is the element added at that step, namely $i_j$. Therefore the label word of the chain is exactly the permutation $(i_1,\dots,i_k)$. This is the key reduction: checking the EL-condition on $[S,T]$ is the same as checking a statement about permutations of the finite set $T \setminus S$.[/guided]

[step:Show that the increasing chain is unique and lexicographically first] Let $(b_1,\dots,b_k)$ be the increasing listing of $T \setminus S$, so \begin{align*} b_1 < b_2 < \cdots < b_k. \end{align*} The corresponding maximal chain is given by \begin{align*} S = S_0 \lessdot S_1 \lessdot \cdots \lessdot S_k = T, \qquad S_j = S \cup \{b_1,\dots,b_j\}. \end{align*} It has strictly increasing label word $(b_1,\dots,b_k)$. If a maximal chain in $[S,T]$ has weakly increasing label word, then its label word is a weakly increasing permutation of the distinct elements of $T \setminus S$. Hence it must equal $(b_1,\dots,b_k)$. Thus the rising maximal chain is unique. Now let $(i_1,\dots,i_k)$ be any other permutation of $T \setminus S$. Let $r$ be the first index such that $i_r \neq b_r$. Since $i_1=b_1,\dots,i_{r-1}=b_{r-1}$, the remaining unused entries in the increasing word begin with the least remaining element $b_r$. Therefore $b_r < i_r$. Hence \begin{align*} (b_1,\dots,b_k) <_{\mathrm{lex}} (i_1,\dots,i_k). \end{align*} So the unique rising maximal chain is lexicographically first among all maximal chains in $[S,T]$. Therefore $\lambda$ is an EL-labelling of $B_n$. [/step]

[step:Shell the proper part by lexicographic order on facets] Let $\overline{B}_n := B_n \setminus \{\varnothing,[n]\}$ be the proper part of $B_n$, and let $\Delta(\overline{B}_n)$ be its order complex. If $n=1$, then $\overline{B}_1 = \varnothing$, so $\Delta(\overline{B}_1)$ is the empty complex, which is shellable by the standard convention. Assume now that $n \geq 2$. For each permutation $\pi = (\pi_1,\dots,\pi_n)$ of $[n]$, define a facet $F_\pi$ of $\Delta(\overline{B}_n)$ by \begin{align*} F_\pi := \{A_1^\pi,\dots,A_{n-1}^\pi\}, \qquad A_r^\pi := \{\pi_1,\dots,\pi_r\} \quad \text{for } 1 \leq r \leq n-1. \end{align*} Every facet of $\Delta(\overline{B}_n)$ has this form, because a maximal chain in $\overline{B}_n$ contains exactly one subset of each cardinality $1,\dots,n-1$. Order the facets lexicographically by the associated words $\pi$. We verify the shelling criterion: for every pair of facets $F_\sigma$ and $F_\pi$ with $F_\sigma$ earlier than $F_\pi$, it suffices to find an earlier facet $F_\tau$ and a vertex $v \in F_\pi \setminus F_\sigma$ such that $F_\pi \setminus F_\tau = \{v\}$. Let $F_\sigma$ and $F_\pi$ be facets with $\sigma <_{\mathrm{lex}} \pi$. Let $m$ be the first index such that $\sigma_m \neq \pi_m$. Then $\sigma_m < \pi_m$. Let $\ell$ be the unique index with $\pi_\ell = \sigma_m$. Since the words agree before $m$, the element $\sigma_m$ does not occur among $\pi_1,\dots,\pi_{m-1}$, so $\ell \geq m$. Since $\pi_m \neq \sigma_m$, in fact $\ell > m$. The finite sequence \begin{align*} \pi_m,\pi_{m+1},\dots,\pi_\ell \end{align*} begins with $\pi_m$ and ends with $\sigma_m$, with $\sigma_m < \pi_m$. Hence it cannot be weakly increasing throughout. Therefore there exists an index $q \in \{m,\dots,\ell-1\}$ such that \begin{align*} \pi_q > \pi_{q+1}. \end{align*} Let $\tau$ be the permutation obtained from $\pi$ by interchanging the adjacent entries $\pi_q$ and $\pi_{q+1}$ and leaving all other entries fixed. Since $\pi_q > \pi_{q+1}$, the first position where $\tau$ and $\pi$ differ is $q$, and $\tau_q = \pi_{q+1} < \pi_q$. Thus \begin{align*} \tau <_{\mathrm{lex}} \pi. \end{align*} The facets $F_\tau$ and $F_\pi$ differ only in rank $q$, because adjacent transposition changes only the $q$-element initial set. More precisely, \begin{align*} A_r^\tau = A_r^\pi \quad \text{for every } r \neq q, \end{align*} while \begin{align*} A_q^\tau = A_{q-1}^\pi \cup \{\pi_{q+1}\} \neq A_q^\pi. \end{align*} Hence \begin{align*} F_\pi \setminus F_\tau = \{A_q^\pi\}. \end{align*} It remains to check that $A_q^\pi \notin F_\sigma$. Since $q \geq m$, the element $\sigma_m$ belongs to $A_q^\sigma$. Since $q < \ell$ and $\pi_\ell=\sigma_m$, the element $\sigma_m$ does not belong to $A_q^\pi$. Therefore $A_q^\pi \neq A_q^\sigma$. Also $|A_q^\pi| = q$, and a chain in the Boolean lattice contains at most one set of cardinality $q$. Thus $A_q^\pi$ is not a vertex of $F_\sigma$. We have shown that for every earlier facet $F_\sigma <_{\mathrm{lex}} F_\pi$, there is an earlier facet $F_\tau <_{\mathrm{lex}} F_\pi$ and a vertex $A_q^\pi \in F_\pi \setminus F_\sigma$ such that \begin{align*} F_\pi \setminus F_\tau = \{A_q^\pi\}. \end{align*} This is the standard facet-by-facet shelling criterion for a pure simplicial complex. Therefore the lexicographic ordering of the facets $F_\pi$ is a shelling of $\Delta(\overline{B}_n)$. Consequently, the order complex of the proper part of $B_n$ is shellable. [/step]