The number of cliques in a graph: the Moon and Moser 1965 result
I don't have a copy of Moon &amp; Moser at hand, but: the maximum number of distinct maximal cliques in an $n$-node graph (with $n>1$) is $3^{n/3}$, $4\cdot 3^{(n-4)/3}$, or $2\cdot 3^{(n-2)/3}$, according to the value of $n$ mod 3. I think it's a little easier to see this in the complementary form of counting maximal independent sets.

The lower bound is what you're really asking for, and is mostly given already in the comments above: form a graph from the disjoint union of copies of $K_2$ and $K_3$, using as many copies of $K_3$ as possible. Each maximal independent set has exactly one node from each of these complete subgraphs from which the formula follows.

I seem to recall that the upper bound proof of Moon and Moser involved transforming a graph into the lower bound form (or the complementary form for maximal cliques), at each step not decreasing the number of independent sets or cliques. But there's a different way of proving it that leads to worst-case-optimal backtracking algorithms for listing all the cliques or independent sets (see e.g. my paper arXiv:cs/0011009), which I'll only sketch here because the details are a bit tedious. If there's a vertex $v$ of degree three or more in the given graph $G$, then each maximal independent set of $G$ is either a maximal independent set in $G\setminus v$ or it includes $v$ and is a maximal independent set of the graph formed from $G$ by removing $v$ and all of its neighbors. By induction (plugging in the formula for the number of independent sets in these two smaller graphs, with some case analysis mod 3) the bound follows. On the other hand, if no high degree vertex exists then the graph is a disjoint union of paths and cycles, in which one can calculate the number of independent sets directly.