Prove that if $R_1, \dots, R_n$ are rings, then the ideals of
$R_1 \times \dots \times R_n$
are all of the form
$\mathcal{I}=\mathcal{I}_1 \times \dots \times \mathcal{I}_n$
where $\mathcal{I}_i \subseteq R_i$ is an ideal. [Hint: consider what happens when you multiply the ideal on the left by $(1,0,\dots,0), (0,1,\dots,0),\dots$.]
Suppose that $\mathcal{I}$ is such that more than one of the $\mathcal{I}_i$'s is different from $R_i$; then we can replace one of these $\mathcal{I}_i$'s with $R_i$ and get an ideal properly containing $\mathcal{I}$. Hence a maximal ideal has all $\mathcal{I}_i$'s but one equal to $R_i$. It is clear that the one $\mathcal{I}_i$ with $\mathcal{I}_i \neq R_i$ is also maximal.
Note: it is no longer true for infinite products of rings. For instance $\prod_{i=1}^\infty \mathbb{Z}/2\mathbb{Z}$ has uncountably many maximal ideals but only countably many maximal ideals of the form just described.