Let $X$ and $Y$ be finite CW complexes. We equip $X\times Y$ with the usual CW complex structure.
Thus $C^{CW}_*(X \times Y) \cong C^{CW}_* (X) \otimes C^{CW}_*(Y)$ as chain complexes.
I know for cellular homology there's an explicit formula for the boundary map $d$ given by $d(D^k_i)= \sum_j n_{i,j} D^{k-1}_j $ where $n_{i,j}$ is the degree of the composite map $S_i^{k-1} \rightarrow X^{(k-1)} \rightarrow X^{(k-1)}/X^{(k-2)} \cong \lor_j S_j^{k-1}\rightarrow S^{k-1}_j$ where the first arrow is the attaching map to $X^{(k-1)}$
For the product $X\times Y$ the formula should be $d(D^i_\alpha \times D^{k-i}_\beta )= d(D^i_\alpha) \otimes D^{k-i}_\beta + (-1)^i D^i_\alpha \otimes d(D^{k-i}_\beta)$
To get this I start with the same recipe. Let's take a $p+q$ cell in $ X\times Y$ say $D^p_i\times D^q_j$. Then i have to look at the attaching map from its boundary $S^{p-1}_i \times D^q_j \cup D^p_{i} \times S^{q-1}_j $ to $(X\times Y)^{(p+q-1)} $
When I quotient this by the $p+q-2$th subcomplex i get $ \lor S^{p+q-1} $ but due to the product nature of attaching the simplices i can identify this with $ \lor ( S^{i-1}_\gamma \times D^{(p+q-1-i )}_\delta \cup D^i _\gamma \times S^ {p+q-i-2} _\delta )$.
Also i can observe (not very clearly though) that $ S^{p-1}_i \times D^q_j $ gets mapped to $ S^{p-1}_\gamma \times D^q_\delta$ after considering the composite. After this point i am stuck since i dont know how that $(-1)^p $ will come by this argument.
Any help/ insight/reference/ corrections will be highly appreciated.