How many vertices can a convex polytope have?

Using $k$ half-spaces, the polytope has at most $k$ facets. For a fixed number of facets, the number of vertices is maximized, for example, by the dual polytope of the cyclic polytope with $k$ vertices. More generally, by the dual polytope of any neighborly polytope with $k$ vertices. This maximum number of vertices is equal to $${k-\lceil n/2 \rceil \choose \lfloor n/2 \rfloor} + {k-\lfloor n/2 \rfloor - 1 \choose \lceil n/2 \rceil - 1}.$$

What you call $n$ in your posting is often called $d$ in the literature, but I will stick with your notation. So $n$ is the dimension. Let $V$ be the number of vertices, and $k$ the number of facets. Then $V = \Theta( k ^ {\lfloor n/2 \rfloor} )$. More precisely, the maximum $V$ is given by McMullen's Upper Bound Theorem, realized by duals of cyclic polytopes. Cyclic polytopes maximize the number of facets for a fixed number of vertices, so their duals maximize the number of vertices for a fixed number of facets.
    
See, e.g.,

"Basic Properties Of Convex Polytopes." Martin Henk, Jürgen Richter-Gebert, and Günter M. Ziegler. Handbook of Discrete and Computational Geometry, Chapter 16. CRC Press. 2004. (Citeseer link)