Free groups are residually finite

By [L1], replace $g$ by a conjugate and assume it is represented by a nontrivial cyclically reduced word $w=x_1\cdots x_n$ in finitely many letters. Build a finite pointed labelled graph with distinct vertices $v_0,\dots ,v_n$ and a distinguished path from $v_{k-1}$ to $v_k$ labelled $x_k$. Because $w$ is reduced, these prescribed edges define partial injective transitions. Complete each partial transition to a permutation of the finite vertex set, so every vertex has exactly one incoming and one outgoing edge of each used label.

Reading words from $v_0$ gives an action of the free group on the finite vertex set. The word $w$ sends $v_0$ to the distinct vertex $v_n$. Therefore the induced homomorphism $F(X)\to \mathrm{Sym}(V)$ does not kill $g$.

The kernel of that finite permutation action has finite index and omits $g$. Since $g\mathrm{\ne }1$ was arbitrary, [F1] shows that the finite residual is trivial. Hence free groups are residually finite.