I am trying to prove that if $T$ is a complete first order theory that has a finite model then it has exactly one model up to isomorphism. To this end, I assumed that $T$ is complete with a finite model $M_n$. Then I assumed that $M_m$ was another model of size $m \geq n$. We know that if a theory has two finite models with different cardinalities then the theory is incomplete hence $m = n$. Now two things remain to be shown: one is that any two models of finite size $n$ are isomorphic and the other is that every infinite model is also isomorphic to this finite model (is this even possible? but we clearly need to do something about the infinite case)
Thanks for helping me finish this proof. For the record: this is an exercise in Just/Weese, page 84.
Edit
I'm looking for a sentence $\varphi$ such that if $M_n \models \varphi$ then $M_n \cong M_m$. There are no assumptions on the language. But I think it's not possible to have a finite model for an infinite language so the language must be finite.
Edit 2
After some more thinking, if there is a finite model $M$ of size $n$, let $ \varphi = \exists v_1, \dots , v_n ((v_1 \neq v_2) \land \dots \land (v_{n-1} \neq v_n)) \land \lnot \exists v_{n+1} ((v_{n+1} \neq v_1) \land \dots \land (v_{n+1} \neq v_n))$ that is, $\varphi$ says that the model has exactly $n$ elements. Since $T$ is complete, either $\varphi$ or $\lnot \varphi$ is provable from $T$. Since we have a model in which $\varphi$ is true we therefore know that $T \vdash \varphi$. Hence any model of $T$ must have exactly $n$ elements.
Now the question is, how do I show that any two $n$-element models of $T$ must be isomorphic?