The prism triangulation has the stated oriented boundary

If $n=0$, then ${\mathrm{\Pi }}_0={\lambda }_0$ is the oriented edge from $(v_0,0)$ to $(v_0,1)$, so [L2] gives $\mathrm{\partial }{\mathrm{\Pi }}_0={\iota }_1-{\iota }_0$. This is exactly the displayed formula with no side-prism sum.

Assume $n\ge 1$. By [L2], each $\mathrm{\partial }{\lambda }_i$ is the alternating sum of its $n+2$ codimension-one faces. The face opposite $(v_0,0)$ in ${\lambda }_0=[(v_0,0),(v_0,1),\dots ,(v_n,1)]$ is the top face ${\iota }_1$, and the face opposite $(v_n,1)$ in ${\lambda }_n=[(v_0,0),\dots ,(v_n,0),(v_n,1)]$ is the bottom face ${\iota }_0$. For each $1\le i\le n$, the face of ${\lambda }_i$ opposite $(v_i,0)$ is the same $n$-simplex as the face of ${\lambda }_{i-1}$ opposite $(v_{i-1},1)$, so these interior faces occur twice in ${\sum }_{i=0}^n(-1{)}^i\mathrm{\partial }{\lambda }_i$ with opposite total signs and cancel.

The remaining uncancelled faces are the top face ${\iota }_1$ from ${\lambda }_0$, the bottom face $-{\iota }_0$ from ${\lambda }_n$, and the side faces obtained by deleting one vertex $v_j$ of ${\mathrm{\Delta }}^n$. For each fixed $j$, those side faces are exactly the prism simplices in the standard triangulation of ${\delta }_j({\mathrm{\Delta }}^{n-1})\times I$, and comparing the induced vertex order with the definition of ${\mathrm{\Pi }}_{n-1}$ gives the total contribution $-(-1{)}^j({\delta }_j\times {\mathrm{id}}_I{)}_{\mathrm{\#}}{\mathrm{\Pi }}_{n-1}$. Summing over $j$ and combining with steps 1.1 and 1.2 yields the stated boundary formula.