Upper bound for #Monotone k-SAT (I've recently started studying satisfiability problems. I've tried to be as clear as possible, but I'm not sure if all of the terminology used is correct.) 

Consider a collection of $n$ Boolean variables, $x_1,\ldots,x_n$ and a function $F$ on this collection of variables, $F(x_1,\ldots,x_n)$. An assignment of true and false values to each Boolean variable is called a satisfying assignment if $F(x_1,\ldots,x_n)$ is true.

I'm interested in functions $F$ that take the form $(x_{i_1}\vee x_{j_1} \vee \cdots \vee x_{k_1})\wedge \cdots \wedge (x_{i_m}\vee x_{j_m} \vee \cdots \vee x_{k_m})$ where indices are allowed to repeat, a so-called Monotone k-SAT function. Perhaps something worth noting is that not all of the clauses contain $k$ variables, but it is known that $k$ is the most number of variables in all clauses. 

Question: What sort of bounds on the number of satisfying assignments for $F$ are known? The worst bound is $2^n$; I'm hoping there is something better.

(Note: this builds on an original question graciously answered by Yuval Filmus here: Number of states in an AND-OR DAG)
A $k$-SAT instance over $n$ variables, monotone or otherwise, has at most $2^n - 2^{n-k}$ solutions.  This is simply the number of possible settings of the $n$ variables minus the least number of settings a single $k$-CNF clause can exclude, assuming all the variables in the clause are distinct.  You only have to consider the case of a formula with a single clause because adding more clauses can never increase the number of satisfying assignments.