ScalingStacks

Proof. For any pair $(D,W)$ of category $D$ and subcategory $W$, we have that $\we(\widetilde{D^{[n]}})=W^{[n]}$, and that $\nerve(W^{[n]})=(\nerve W)^{\Delta[n]}$. We obtain the result by substituting $C^{[m]}$ for $D$. ∎