Boundary map for product of CW complexes 
why the boundary operator in the cellular product complex satisfies the "product rule"?

For some references, I was reading this pdf, and in Exercise 3, it is written that the boundary operator satisfies the relation $$ d(e_{\alpha}^m\times e_{\beta}^n) = d(e_{\alpha}^m)\times e_{\beta}^n + (-1)^m e_{\alpha}^m \times d(e_{\beta}^n)$$
but in that notes there is not any hints on why this has to hold.
 A: You can look at it in geometric way. Think of a pair of manifolds with boundary $(M,N)$, for example $(I,I)$, $(I, D^2)$, $(D^2, D^2)$, ... 
Which are the boundary points of the product manifold $M\times N$? Intuitively, a point $(m,n)\in M\times N$ should be in $\partial(M\times N)$ if $m\in \partial M$ or $n\in\partial N$. Check this for some examples: the square $I\times I$, the cylinder $I\times D^2$, ...
Hence, the topological boundary would be
$$
\partial (M\times N) = (\partial M)\times N \cup M \times (\partial N).
$$
Now in algebraic topology boundaries of cells of CW-complexes are algebraic versions of the topological boundaries: linear combinations of the cells contributing to the boundary with coefficients chosen two reflect the attaching maps of cells and signs chosen such that $d\circ d=0$.
This should give you an intuition for why $d(e_\alpha^m\times e_\beta^n)$ is a linear combination of the two chains $d(e_\alpha^m)\times e_\beta^n$ and $e_\alpha^m \times d(e_\beta^n)$.
If you want to derive this from the cell decomposition of products of CW-complexes, you'd have to look at the degrees of attaching maps in the product complex.
