Size comparison of NFA and minimal DFA
For every $n$ there is a language over $\{a,b\}$ which is accepted by an NFA having $n$ states, but its minimal DFA has $2^n$ states.

Unfortunately I don't remember the reference, but here is how to get almost this result. Let $L_n$ be the language of all words over $\{a_1,\ldots,a_n\}$ that don't contain all letters. Then $L_n$ can be accepted by an NFA with multiple initial states having $n$ states (or with a normal NFA with $n+1$ states), but its minimal DFA has $2^n$ states. Take it as an exercise to prove these statements (for the latter one, use Myhill-Nerode theory).