associativity in graph theory

Can anybody help me in clearing the facts how the associativity was proved in cartesian product of 3 graphs, and thus showing isomorphism. I can easily solve for the case when its two graphs. Taking three graphs is not clear to me. Kindly help. Thanks a lot to all.

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It's sometimes beneficial to think about the Cartesian product of graphs in analogy to the ordinary Cartesian product. For example, product of two lines gives us a plane, product of two discrete lines results in a discrete plane. Similarly, if we arrange the vertices of the graph into a finite and discrete line, then what we will get would look like finite discrete plane (with appropriate edges). Also, as in $\mathbb{N^2}$, you can go north and south, and east and west, but you cannot use diagonals, in similar fashion you can use $\color{blue}{\text{blue}}$ edges to go east and west and you can use $\color{green}{\text{green}}$ edges to go north and south, but there are no diagonal edges.
When you add a third graph, things doesn't change much. As you can produce $\mathbb{N^3}$ as $\mathbb{N}^2 \times \mathbb{N}$ or $\mathbb{N} \times \mathbb{N}^2$, the final structure doesn't change. This is because each graph allows you to move in some "dimension", but nothing more, there are no diagonals (which might make the product non-associative). If you want to go up or down, you have to use $\color{red}{\text{red}}$ edges no matter whether the "base graph" is, $\color{blue}{\text{blue}} \times \color{green}{\text{green}}$ or just $\color{green}{\text{green}}$. Whatever order you would take the product of graphs, the only edges are $\color{blue}{\text{blue}}$ to move between $(\color{blue}{x_1},y,z)$ and $(\color{blue}{x_2},y,z)$, $\color{green}{\text{green}}$ edges to move between $(x, \color{green}{y_1},z)$ and $(x,\color{green}{y_2},z)$ and finally $\color{red}{\text{red}}$ edges to move between vertices $(x, y, \color{red}{z_1})$ and $(x,y, \color{red}{z_2})$. Moreover, whatever order you would take the product of graphs, you will get all those edges, and as such, the graphs will be isomorphic.
I hope this helps $\ddot\smile$