From Rotman's Algebraic Topology:
Let $S_{n}(X) = \{\sigma | \sigma : \Delta^n \rightarrow X\}$ be the free abelian group with basis all singular $n$-simplices $\sigma: \Delta^n \rightarrow X$.
For all $n$, $\partial_n \partial_{n+1} = 0$.
Proof:
$\partial \partial \sigma = \partial(\sum_j (-1)^j\sigma \epsilon_j^{n+1}) = \sum _{j,k}(-1)^{j+k}\sigma\epsilon_j^{n+1}\epsilon_k^{n} =$
$ \sum _{j \le k}(-1)^{j+k}\sigma \epsilon_j^{n+1}\epsilon_k^n + \sum _{k \lt j}(-1)^{j+k}\sigma \epsilon_k^{n+1}\epsilon_{j-1}^n$ .
In the second sum, change the variables to $p = k$ and $q = j-1$ and now each term appears twice:
$ \sum _{j \le k}(-1)^{j+k}\sigma \epsilon_j^{n+1}\epsilon_k^n + \sum _{p \le q}(-1)^{p+q+1}\sigma \epsilon_p^{n+1}\epsilon_{q}^n$,
once in the first sum with positive sign and once in the second sum with opposite sign.
How does each term appear twice? Let $p - 2$ and $q = 4$ which implies $k = 2$ and $j = 5$, but $5$ is not less than $2$ and therefore cannot be in the first sum.
What am I missing here?
Notice in the second line, the sum ranges over all pairs of $j,k$ as those indices range over all the subscripts of the edges. Think of this as a big square of summands.
The sums in the third and fourth lines might appear to be sums only along one row or one column of that square (i.e., for fixed $j$ or for fixed $k$), but they are not. They are sums of the upper and lower triangular regions of that square. What is being observed is that for each term in either of these triangles of summands, there is a cancelling summand in the other triangle.
The reason for changing variables is that it is not the case that the $j,k$ term cancels the $k,j$ term.