The simplicial boundary of a boundary is zero

If $k=1$, then ${\mathrm{\partial }}_{k-1}={\mathrm{\partial }}_0=0$ by the definition, so the double boundary vanishes. Now let $k\ge 2$ and fix an oriented generator $[v_0,\dots ,v_k]$. In ${\mathrm{\partial }}_{k-1}{\mathrm{\partial }}_k[v_0,\dots ,v_k]$, a codimension-two face is obtained by deleting a unique pair of vertices with indices $i<j$. Deleting $i$ first contributes coefficient $(-1{)}^i(-1{)}^{j-1}$ because the original index $j$ shifts down by one; deleting $j$ first contributes $(-1{)}^j(-1{)}^i$.

The sum of those two coefficients is $(-1{)}^{i+j-1}+(-1{)}^{i+j}=0$. Every codimension-two face arises from exactly these two deletion orders, so every basis coefficient in the double boundary is zero. The double boundary therefore vanishes on each generator and, by linearity, on all of $C_k$.