# Finite languages and their models

I realize this may look like a duplicate, but I still cannot figure this out.

If we have a finite language, say $\mathcal{L}$, and some set of sentences in it, say $\Gamma$, and $\Gamma$ has no infinite model, then why is it the case that there cannot be infinitely many models for $\Gamma$?

Assume there are infinitely many non isomorphic models of $\Gamma$. Then as for fixed $n$ there is only a finite number of (up to isomorphism) $\mathcal{L}$-structures of size $n$, and so only finitely many (up to isomorphism) models of $\Gamma$ of size $n$.
Therefore, our assumption implies that $\Gamma$ has arbitrarily large models. Let $\mathcal{L'}$ be the language consisting of $\mathcal{L}$ and of distinct constant symbols $\{c_i, i\in \mathbb{N}\}$. Then let $\Gamma' =\Gamma \cup\{c_i \neq c_j, i\neq j\}$.
The fact that $\Gamma$ has arbitrarily large models implies (can you show it ?) that $\Gamma'$ is finitely consistant, and so by compactness, it is consistant and thus has a model. But a model of $\Gamma'$ has to be infinite, and its $\mathcal{L}$-reduct is an infinite model of $\Gamma$.
So by contraposition, if $\Gamma$ has no infinite model, then it has (up to isomorphism) only a finite number of models