$\sum_{i=0}^n (-1)^i \delta_i(\sum_{j=0}^{n+1} (-1)^j \delta_j) = 0$, given that $\delta_i \delta_j = \delta_{j-1}\delta_i$ whenever $i < j$
This problem shows up in the middle of dealing with simplicial sets and homology groups, but it's more of a pure summation-fixing sort of problem, since only the rule given is needed to prove the result.
I know it's true, and I think I could, at least in theory, do it for any n. For example: $n=2: \sum_{i=0}^2 (-1)^i \delta_i(\sum_{j=0}^{3} (-1)^j \delta_j)$ $=\delta_0 \delta_0 - \delta_0 \delta_1 + \delta_0 \delta_2 - \delta_0 \delta_3 -\delta_1 \delta_0 + \delta_1 \delta_1 - \delta_1 \delta_2 + \delta_1 \delta_3 + \delta_2 \delta_0 - \delta_2 \delta_1 + \delta_2 \delta_2 - \delta_2 \delta_3$ $=\delta_0 \delta_0 - \delta_0 \delta_0 + \delta_1 \delta_0 - \delta_2 \delta_0 -\delta_1 \delta_0 + \delta_1 \delta_1 - \delta_1 \delta_1 + \delta_2 \delta_1 + \delta_2 \delta_0 - \delta_2 \delta_1 + \delta_2 \delta_2 - \delta_2 \delta_2$ $=\delta_1 \delta_0 - \delta_2 \delta_0 - \delta_1 \delta_0 + \delta_2 \delta_1 + \delta_2 \delta_0 - \delta_2 \delta_1$ $=0$.
Now, when I try to do it in general, using summation-notation, this is what I get: $\sum_{i=0}^n (-1)^i \delta_i(\sum_{j=0}^{n+1} (-1)^j \delta_j)$ $= \sum_{i=0}^n \sum_{j=0}^{n+1} (-1)^{i+j} \delta_i \delta_j$ $= \sum_{j=0}^{n+1} \sum_{i=0}^n (-1)^{i+j} \delta_i \delta_j$ $= \sum_{i=0}^n (-1)^i \delta_i \delta_0 + \sum_{j=1}^{n+1} (\sum_{i=0}^{j-1} (-1)^{i+j} \delta_i \delta_j + \sum_{i=j}^n (-1)^{i+j} \delta_i \delta_j)$ $= \sum_{i=0}^n (-1)^i \delta_i \delta_0 + \sum_{j=1}^{n+1} (\sum_{i=0}^{j-1} (-1)^{i+j} \delta_{j-1} \delta_i + \sum_{i=j}^n (-1)^{i+j} \delta_i \delta_j)$ $= \sum_{i=0}^n (-1)^i \delta_i \delta_0 + \sum_{j=0}^{n} (\sum_{i=0}^{j} (-1)^{i+j+1} \delta_{j} \delta_i + \sum_{i=j+1}^n (-1)^{i+j+1} \delta_i \delta_{j+1})$.
I'm not sure if that last step is necessary/wise/correctly done, but whether or not it is, the trouble now is I'm not quite sure how to shunt around the terms so that the matching pairs are obvious enough to remove.
Here you go, I will write $d_i$ instead of $\delta_i$ (because I'm lazy): $$\begin{align} \sum_{i=0}^n \sum_{j = 0}^{n+1} (-1)^{i+j} d_i d_j & = \sum_{0 \le i < j \le n+1} (-1)^{i+j} d_i d_j + \sum_{0 \le j \le i \le n} (-1)^{i+j} d_i d_j \\ & = \sum_{0 \le i < j \le n+1} (-1)^{i+j} d_{j-1} d_i + \sum_{0 \le j \le i \le n} (-1)^{i+j} d_i d_j = \dots \\ \end{align}$$ Now in the first sum I set $a = j-1$ and $b = i$, while in the second sum I set $a=i$, $b=j$: $$\dots = \sum_{0 \le b \le a \le n} (-1)^{a+b-1} d_{a} d_{b} + \sum_{0 \le b \le a \le n} (-1)^{a+b} d_a d_b = 0.$$