Do a set of vertices uniquely determine a polytope? The question in title arises because I am trying to prove that a polytope is the convex hull of its vertices, i.e., $\mathcal{P}=conv(V)$. Here is how far I have got.
Convex hull of a finite set of vectors is a polytope. So for $v_1,...,v_k$, $\mathcal{Q}=conv(v_1,...v_k)$ is a polytope and I can show that for this polytope, $\mathcal{Q}$, $v_i$ must be vertices for all $i$. So I now know that if I take convex hull of a set of vectors, I get a polytope whose vertices are those vectors. But I require that vertices determine a polytope uniquely or else I cannot complete my proof. Is this true and if not, how do I prove that a polytope is convex hull of its vertices?

Yes, in general this is how one defines a convex polytope. In general, a convex polytope is defined as the convex hull of a finite set (of vertices), page 14:

Def: A convex polytope $P\subset \mathbb{R}^n$, or symply a polytope, is defined as the convex hull of a non empty finite set $\{x_1,\dots,x_q\}$

So maybe you should provide the definition of convex polytopes that you are using. There another way to characterize a convex polytope $P$. Which is by means of a finite set of inequalities. Some times this is refered as the H-V theorem were H stands for half-spaces and V for vertices. Sometimes also as the Fundamental Theorem of Polytopes:

Theorem 9.2: A non-empty set $P$ of $\mathbb{R}^n$ is a (convex) polytope if, and only if, it is a bounded polyhedral set.

Polyhedral set is defined in the same text as

Def: A subset $Q$ of $\mathbb{R}^n$ is called a polyhedral set if $Q$ is the intersection of a finite number of closed halfspaces, or $Q=\mathbb{R}^n$.

