Every DFA language is denoted by a regular expression

Form a GNFA $G_0$ by adjoining a new start state $s$ and a new accept state $t$ to the DFA states. Put the label $\epsilon$ on $s\to q_0$ and on each edge $f\to t$ with $f\in F$. For DFA states $p,q\in Q$, label $p\to q$ by the union of all letters $a\in \mathrm{\Sigma }$ with $\delta (p,a)=q$, and use $\mathrm{\varnothing }$ on every remaining ordered pair. By [L4], a word is accepted by $D$ exactly when it labels a path from $s$ to $t$ through the old DFA states, so $L(D)=P_Q^{G_0}(s,t)$.

Eliminate the states of $Q$ one at a time. At each step use the update from [L2]. By [L3], after each elimination the start-to-accept path language is unchanged. Therefore, when all interior states have been removed, the resulting two-state GNFA $G_f$ still satisfies $L(G_f)=L(D)$.

In $G_f$ the only possible accepting paths go directly from $s$ to $t$, so by [L1] the accepted language is exactly the denotation of the sole label on that edge. Hence some regular expression denotes $L(D)$.