Proving that no finite automaton can recognize a Dyck language
Consider a Dyck language $L$ with $k$ types of parentheses, one of them $()$. If $L$ were regular then $L \cap (^*)^*$ would be regular, but this is just the well-known non-regular language $a^nb^n$ (with $a,b$ replaced by $(,)$).

There are many ways to prove that $L' = \{a^nb^n : n \geq 0\}$ isn't regular. Here is one. Suppose that $A$ is a DFA recognizing $L'$, with states $Q$, initial state $q_0$, accepting states $F$, and transition function $\delta$. Let $q(i) = \delta(q_0,a^i)$. If $i &lt; j$ then $q(i) \neq q(j)$, since otherwise $\delta(q_0,a^ib^i) = \delta(q_0,a^jb_i)$ although the former is in $F$ while the latter isn't. This implies the absurd conclusion that $Q$ must be infinite.