For \(k=2\) we have

\[

\partial \sigma=\left[\mathbf{p}_{1}, \mathbf{p}_{2}\right]-\left[\mathbf{p}_{0}, \mathbf{p}_{2}\right]+\left[\mathbf{p}_{0}, \mathbf{p}_{1}\right]

\]

so that

\[

\partial^{2} \sigma=\left(\mathbf{p}_{2}-\mathbf{p}_{1}\right)-\left(\mathbf{p}_{2}-\mathbf{p}_{0}\right)+\left(\mathbf{p}_{1}-\mathbf{p}_{0}\right)=0

\]

For \(k=3\) we have

\[

\partial \sigma=\left[\mathbf{p}_{1}, \mathbf{p}_{2}, \mathbf{p}_{3}\right]-\left[\mathbf{p}_{0}, \mathbf{p}_{2}, \mathbf{p}_{3}\right]+\left[\mathbf{p}_{0}, \mathbf{p}_{1}, \mathbf{p}_{3}\right]-\left[\mathbf{p}_{0}, \mathbf{p}_{1}, \mathbf{p}_{2}\right]

\]

so that

\[

\begin{aligned}

\partial^{2} \sigma=&\left(\left[\mathbf{p}_{2}, \mathbf{p}_{3}\right]-\left[\mathbf{p}_{1}, \mathbf{p}_{3}\right]+\left[\mathbf{p}_{1}, \mathbf{p}_{2}\right]\right) \\

&-\left(\left[\mathbf{p}_{2}, \mathbf{p}_{3}\right]-\left[\mathbf{p}_{0}, \mathbf{p}_{3}\right]+\left[\mathbf{p}_{0}, \mathbf{p}_{2}\right]\right) \\

&+\left(\left[\mathbf{p}_{1}, \mathbf{p}_{3}\right]-\left[\mathbf{p}_{0}, \mathbf{p}_{3}\right]+\left[\mathbf{p}_{0}, \mathbf{p}_{1}\right]\right) \\

&-\left(\left[\mathbf{p}_{1}, \mathbf{p}_{2}\right]-\left[\mathbf{p}_{0}, \mathbf{p}_{2}\right]+\left[\mathbf{p}_{0}, \mathbf{p}_{1}\right]\right) \\

=& 0

\end{aligned}

\]

In general the order in which \(\mathbf{p}_{i}\) and \(\mathbf{p}_{j}\) are omitted from \(\sigma\) determines the sign that \(\sigma_{i j}\) will have. If \(\mathbf{p}_{j}\) is omitted first, the resulting \((k-1)\)-simplex \(\sigma_{j}=\) \(\left[\mathbf{p}_{0}, \ldots, \mathbf{p}_{j-1}, \mathbf{p}_{j+1}, \ldots, \mathbf{p}_{k}\right]\) will acquire the sign \((-1)^{j}\). If \(\mathbf{p}_{i}\) is then omitted, the resulting \((k-2)\)-simplex will acquire a factor of \((-1)^{i}\), resulting in \((-1)^{i+j} \sigma_{i j}\).

However, if \(\mathbf{p}_{i}\) is omitted first, \(\mathbf{p}_{j}\) will move forward one position in the resulting \((k-1)\)-simplex \(\sigma_{i}\), and when it is subsequently omitted, a factor of

\((-1)^{j-1}\) will be affixed, resulting in \((-1)^{i+j-1} \sigma_{i j}\). Hence the two occurrences of \(\sigma_{i j}\) in the second boundary will cancel each other.

The linearity of the boundary operator, operating on a base of simplexes, then shows that \(\partial^{2}\) is the zero operator on all chains.