Proof: the number of maximal cliques in a graph with boxicity $d$ is $\mathcal{O}((2n)^d)$ I'm looking for the proof of the following result

The number of maximal cliques in a graph with boxicity $d$ is $\mathcal{O}((2n)^d)$

On the Wikipedia page and other papers where this result has been mentioned, they cite the following paper

Roberts, Fred S. "On the boxicity and cubicity of a graph." Recent progress in combinatorics 1 (1969): 301-310.

Unfortunately, I haven't been able to find this paper anywhere. I'm looking for either access to this paper or a proof sketch for this result. I'm hoping to use the ideas within this proof to improve the bound for a subset of graphs with boxicity $d$.
 A: We actually do a bit better and get a hard bound of $n^d$, not $O((2n)^d)$. Of course, for fixed $d$ as $n \to \infty$, $O(n^d)$ and $O((2n)^d)$ are the same...
The maximal cliques are obtained by picking a point $x \in \mathbb R^d$ and taking all boxes containing $x$. Though there seem to be infinitely many ways to do this, many of them give the same clique - without loss of generality, we may assume that each component $x_i$ of $x$ is the lower bound of the $i^{\text{th}}$ dimension of some box.
To prove this: suppose that a clique is formed by vertices $v_1, v_2, \dots, v_k$, where $v_j$ is represented by the box $[a_1^j, b_1^j] \times \dots \times [a_d^j, b_d^j]$. Let $x \in \mathbb R^d$ be given by $x_i = \max\{a_i^1, \dots, a_i^k\}$. Then:

*

*in the $i^{\text{th}}$ dimension, $x_i \ge a_i^j$ for all $j$, by construction. In particular, $x_i = a_i^{j^*}$ for some $j^*$.

*in the $i^{\text{th}}$ dimension, $x_i \le b_i^j$ for all $j$; otherwise, if $x_i = a_i^{j^*} > b_i^j$, then $[a_i^j, b_i^j]$ and $[a_i^{j^*}, b_i^{j^*}]$ don't intersect, so $v_j$'s and $v_{j^*}$'s boxes don't intersect.

*Therefore $x_i \in [a_i^j, b_i^j]$ for all $j$, which means $x$ is contained in all $k$ boxes.

So the clique given by $v_1, \dots, v_k$ is contained in the clique of all boxes containing $x$; if the clique was maximal, it must be the clique of all boxes containing $x$.
There are $n^d$ ways to choose a point whose every component is the lower bound of some box. Not all of these $n^d$ points will give maximal cliques (some might not even be contained in any boxes) but all of the maximal cliques will be given by one of these points, so there are at most $n^d$ maximal cliques.
