Logic: Can you drop parentheses in a conjunction? In propositional logic, $p \land (q \land r) = (p \land q) \land r$ , where $p, q$ and $r$ are propositions.

Does this mean $p \land (q \land r) = p \land q \land r$ ? If so, why?
Yes, you are right: conjunction is associative. Thus, it doesn't matter where you put the parentheses. Hence, when you see "$p \land q \land r$", it means that you may insert the parentheses anywhere and you will get the same result.
Informally, one might say that if three things are true ($p \land q \land r$), then also one thing and two things &mdash; $p \land (q \land r)$ &mdash; and two things and one thing &mdash; $(p \land q) \land r$ &mdash; are true.

In a formal deductive system commonplace in symbolic logic, we want to be more formal. We usually deduce consequences from a set of premises. For example we deduce $p \land q$ from the set $\{p, q\}$. This is written:
$$\{p,q\} \vdash p \land q$$
where the braces are usually omitted, and $\vdash$ is voiced proves. Similarly:
$$p \land q \vdash p \qquad\qquad p\land q \vdash q$$
(where the premises thus are considered singleton sets) because this is simply throwing information away. 

If we now allow chains of deduction, then, one easily obtains:
$$\begin{align*}\{p,q,r\} &amp;\vdash p \land q&amp; \{p \land q, r\} &amp;\vdash (p \land q) \land r\\
\{p,q,r\}&amp;\vdash q \land r&amp;\{p, q \land r\}&amp;\vdash p \land(q \land r)
\end{align*}$$
and from the rightmost conclusion the so-called $\land$-elimination rules  easily allow deriving $p, q$ and $r$ again. Thus either of the parenthesized expressions corresponds exactly to the truth of the three letters $p,q,r$. Thus we say they're (provably) equivalent:
$$p \land (q \land r) \dashv\vdash (p \land q) \land r$$

Thus we have demonstrated that the following definitions of $p \land q \land r$ are all equally sensible:


As $p \land (q \land r)$;
As $(p \land q) \land r$;
By stating $p \land q \land r$ is true precisely when $p,q,r$ are all true.