The subset construction can require exponentially many states

Let $L_n$ be the set of words over ${\mathrm{\Sigma }}_n$ for which at least one letter $a_i$ occurs an odd number of times. An NFA for $L_n$ has one start state and, for each $i$, two parity states that track whether the number of $a_i$'s seen so far is even or odd. By [L1], the machine may nondeterministically choose which letter $a_i$ to track, and then it accepts exactly when that tracked count ends odd. Thus $L_n$ has an NFA with $1+2n$ states.

For a prefix $u$, let $\pi (u)\in \{0,1{\}}^n$ be its parity vector, whose $i$th coordinate records whether $a_i$ has appeared an even or odd number of times in $u$. Distinct parity vectors are distinguishable: if $\pi (u)\mathrm{\ne }\pi (v)$, append the suffix consisting of exactly those letters whose parity is odd in $\pi (u)$. The resulting word from $u$ has all-even parity and is not in $L_n$, while the resulting word from $v$ still has at least one odd coordinate and lies in $L_n$.

Therefore no DFA can place two different parity vectors in the same state, because [L2] would then force the same acceptance behavior for the distinguishing suffix from step 1.2. Since there are $2^n$ parity vectors, every DFA for $L_n$ needs at least $2^n$ states.

So the subset-construction simulation of NFA's by DFA's can require exponentially many states.