I am trying to prove that if $$d_r (e_{i_1} \wedge e_{i_2} \wedge ... \wedge e_{i_r}) = \sum\limits_{j=1}^r (-1)^{j-1}a_{i_j}(e_{i_1} \wedge e_{i_2} ... \wedge e_{i_(j-1)} \wedge e_{i_(j+1)} \wedge ... \wedge e_{i_r}),$$ then $d_{r-1} \circ d_r = 0$
I worked it out the point where I have $$2 *\sum\limits_{j=1}^r \sum\limits_{k < j} (-1)^{j+k} a_{i_j}a_{i_k} (e_{i_1} \wedge e_{i_2} ... \wedge e_{i_(j-1)} \wedge e_{i_(j+1)} \wedge ... e_{i_{k-1}} \wedge e_{i_{k+1}} \wedge ... \wedge e_{i_r})$$ since there is some symmetry in the way we took the sum. However, I am not sure how I can simplify this to $0$. I know a wedge product is $0$ if there are two terms that are the same.
You had an off by one error. If you think about it the only way this could possibly work is if $$a_{i_j}a_{i_k}\bigwedge_{l=1,l\neq j,k}^r e_{i_l}$$ occurred both positively and negatively.
You should be able to convince yourself that the sum you want is $$\sum_{j = 1}^r\sum_{k=1,j\neq k}^r \varepsilon_{j,k} a_{i_j} a_{i_k}\bigwedge_{l=1,l\neq j,k}^r e_{i_l}$$ where $\varepsilon_{j,k}$ is the sign (i.e. $\varepsilon_{j,k} = \pm 1$). You can then show that $\varepsilon_{j,k}=-\varepsilon_{k,j}$. By symmetry, we can consider just the $j<k$ case, in which case $\varepsilon_{j,k}$ corresponds to $j-1$ followed by $k-\mathbf{2}$ alternations giving $\varepsilon_{j,k}=(-1)^{(j-1)+(k-2)} = (-1)^{j+k-1}$ while $\varepsilon_{k,j}$ corresponds to $k-1$ alternations followed by $j-1$ alternations giving $\varepsilon_{k,j} = (-1)^{(j-1)+(k-1)}=(-1)^{j+k}$. This should be enough to spell this out into a complete proof.