[proofplan]
We identify a face of the deleted join with a partial labelling of the vertices of the simplex by labels in $[r]$. Since $K=\Delta^{n-1}$ is the full simplex, the only restriction is that each original vertex receives at most one label. This gives a simplicial isomorphism from $K_{\Delta}^{*r}$ to the $n$-fold join of the nonempty $0$-dimensional complex $E_r$ consisting of $r$ isolated vertices. The homological connectivity of joins then raises connectivity by one at each join factor, giving homological $(n-2)$-connectedness.
[/proofplan]
[step:Model the deleted join as partial labelings of the simplex vertices]
Let $E_r$ denote the $0$-dimensional simplicial complex with vertex set $[r]=\{1,\dots,r\}$ and with faces precisely $\varnothing$ and the singleton subsets $\{j\}$ for $j \in [r]$.
The vertex set of the ordinary $r$-fold join $K^{*r}$ may be written as $[n]\times [r]$, where $(a,j)$ denotes the vertex $a \in [n]$ placed in the $j$-th join factor. A face of the pairwise deleted join $K_{\Delta}^{*r}$ is a subset $S \subset [n]\times [r]$ such that, for each $j \in [r]$, the set
\begin{align*}
\sigma_j(S) := \{a \in [n] : (a,j)\in S\}
\end{align*}
is a face of $K$, and the sets $\sigma_1(S),\dots,\sigma_r(S)$ are pairwise disjoint.
Because $K=\Delta^{n-1}$ is the full simplex on $[n]$, every subset of $[n]$ is a face of $K$. Hence the only remaining condition is pairwise disjointness of the sets $\sigma_j(S)$. Equivalently, for each $a \in [n]$, there is at most one label $j \in [r]$ such that $(a,j)\in S$. Thus a face of $K_{\Delta}^{*r}$ is precisely a partial function from $[n]$ to $[r]$, encoded by its graph in $[n]\times [r]$.
[guided]
We first unpack the deleted join in this special case. The ordinary $r$-fold join $K^{*r}$ has one copy of $K$ for each label $j \in [r]$. Its vertices are therefore pairs $(a,j)$, where $a \in [n]$ is an original vertex of the simplex and $j$ records the copy of $K$.
A face $S \subset [n]\times [r]$ determines, for every label $j \in [r]$, a subset
\begin{align*}
\sigma_j(S) := \{a \in [n] : (a,j)\in S\}.
\end{align*}
The condition that $S$ is a face of the ordinary join says that each $\sigma_j(S)$ must be a face of $K$. The deleted condition says more: the subsets $\sigma_1(S),\dots,\sigma_r(S)$ must be pairwise disjoint.
Now we use the special hypothesis $K=\Delta^{n-1}$. Since $\Delta^{n-1}$ is the full simplex on $[n]$, every subset of $[n]$ is a face. Therefore the simplex condition imposes no restriction on the sets $\sigma_j(S)$. The only condition left is disjointness. This means that the same original vertex $a \in [n]$ cannot appear in two different labelled copies. In other words, for each $a \in [n]$, there is at most one label $j \in [r]$ with $(a,j)\in S$.
Thus a face of $K_{\Delta}^{*r}$ is exactly a partial labelling of the set $[n]$: some vertices of $[n]$ may be unlabelled, and each labelled vertex receives exactly one label from $[r]$.
[/guided]
[/step]
[step:Identify the partial labelling complex with an $n$-fold join of $E_r$]
For each $a \in [n]$, let $E_{r,a}$ be a copy of $E_r$ whose vertex set is $\{(a,j):j\in [r]\}$. The join
\begin{align*}
J := E_{r,1} * E_{r,2} * \cdots * E_{r,n}
\end{align*}
has vertex set $[n]\times [r]$, and a subset $S \subset [n]\times [r]$ is a face of $J$ exactly when, for each $a \in [n]$, the intersection
\begin{align*}
S \cap \bigl(\{a\}\times [r]\bigr)
\end{align*}
is a face of $E_{r,a}$.
Since the only nonempty faces of $E_{r,a}$ are singletons, this condition is equivalent to saying that, for each $a \in [n]$, the set $S$ contains at most one vertex of the form $(a,j)$. This is exactly the face condition for $K_{\Delta}^{*r}$ obtained in the previous step. Therefore the identity map on the common vertex set $[n]\times [r]$ induces a simplicial isomorphism
\begin{align*}
K_{\Delta}^{*r} \cong E_{r,1} * E_{r,2} * \cdots * E_{r,n}.
\end{align*}
[/step]
[step:Apply homological join connectivity to the $n$ nonempty factors]
Each complex $E_{r,a}$ is nonempty because $r \geq 2$, so
\begin{align*}
\widetilde{H}_{-1}(E_{r,a};k)=0
\end{align*}
for every $a \in [n]$. Thus each $E_{r,a}$ is homologically $(-1)$-connected over $k$.
We use the homological connectivity theorem for joins: if simplicial complexes $A$ and $B$ are homologically $p$-connected and $q$-connected over a field $k$, respectively, then $A*B$ is homologically $(p+q+2)$-connected over $k$ (citing a result not yet in the wiki: Homological Connectivity of Joins). Applying this theorem inductively to the factors $E_{r,1},\dots,E_{r,n}$ gives that
\begin{align*}
E_{r,1} * E_{r,2} * \cdots * E_{r,n}
\end{align*}
is homologically $(n-2)$-connected over $k$.
Indeed, after one factor the connectivity is $-1$. If the join of the first $m$ factors is homologically $(m-2)$-connected, then joining with $E_{r,m+1}$, which is homologically $(-1)$-connected, gives homological connectivity
\begin{align*}
(m-2)+(-1)+2 = m-1.
\end{align*}
This is exactly $(m+1)-2$, so induction proves the claim for all $m=1,\dots,n$.
[/step]
[step:Transfer the homology vanishing through the simplicial isomorphism]
By the simplicial isomorphism constructed above,
\begin{align*}
K_{\Delta}^{*r} \cong E_{r,1} * E_{r,2} * \cdots * E_{r,n}.
\end{align*}
Simplicial isomorphisms induce isomorphisms on reduced homology over $k$. Since the right-hand side is homologically $(n-2)$-connected over $k$, we obtain
\begin{align*}
\widetilde{H}_q(K_{\Delta}^{*r};k)=0
\end{align*}
for every integer $q$ with $-1 \leq q \leq n-2$. This is precisely the asserted homological $(n-2)$-connectedness.
[/step]
