TITLE: Are quantifiers only required because of infinite proposition chains?
QUESTION [5 upvotes]: Quantifiers such as $\forall x \ P(x)$ and $\exists x \ P(x)$ are in some ways equivalent to a long conjunction chain being true versus at least one statement being true in a long disjunction chain. 
Why do we even need quantifiers at all if we can achieve the equivalent thing with conjunctive/disjunctive chains? Is the only reason (besides convenience) that we may have an infinitely long chain?
What does first order logic / quantifiers do that we absolutely cannot do propositional logic?

REPLY [3 votes]: The second part of Bram28's answer I think gets to the real heart of the matter, so let me expand on that.
One of the key purposes of logic is to categorize structures. Any set $\Sigma$ of sentences in a language $L$ defines a corresponding class of $L$-structures: $$Mod(\Sigma)=\{\mathcal{M}: \mathcal{M}\models\Sigma\}.$$ For example, taking $L$ to be the language with a single binary function symbol $*$ and $\Sigma=\{\forall x,y,z((x*y)*z=x*(y*z))\}$ we get that $Mod(\Sigma)$ is the class of semigroups. Some natural classes of structures are first-order characterizable (like the class of semigroups) while others aren't (like the class of torsion groups).
The key point here is that we're looking at the same sentences across different domains. Replacing "$\forall x$" with "$\bigwedge_{x\in dom(\mathcal{M})}$" works in the context of a single fixe structure $\mathcal{M}$, but what if you want to talk about multiple structures at once?
This is especially important if you want to build a structure with certain properties, where I can say "I want my structure to satisfy $\forall x\exists y(P(x,y))$" without having fully determined the domain yet.