The free group on one generator is isomorphic to $(\mathbb Z,+)$

A reduced word on $\{x,x^{-1}\}$ cannot contain both letters, since a change from one to the other creates an adjacent inverse pair; hence every reduced word is uniquely $x^n$ for $n\geq0$ or $x^{-n}$ for $n\geq1$, with the empty word corresponding to exponent $0$.

Thus $[x]$ generates the group, and no positive power $[x]^n$ is the identity because its reduced representative $x^n$ is nonempty; therefore $[x]$ has infinite order.

Since $[x]$ generates by step 2.1, [L5] makes every element of $F_{\mathrm{word}}(\{x\})$ equal to $[x]^k$ for some $k\in\mathbb Z$, and [L2] applied to the infinite-order element $[x]$ makes that exponent unique; so $\theta([x]^k):=k$ is a well-defined injection, it is surjective because $k=\theta([x]^k)$, and [L6] makes it a homomorphism. Construct $\theta:F_{\mathrm{word}}(\{x\})\to(\mathbb Z,+)$ as this isomorphism, which carries $[x]$ to $1$.

By [L4] the word-quotient model is a free group on $X$, so for any free group $(F,i)$ on $X$ the isomorphism $\phi:F\to F_{\mathrm{word}}(\{x\})$ of [L3] satisfies $\phi(i(x))=[x]$; then $\theta\circ\phi$ is an isomorphism $F\to(\mathbb Z,+)$ carrying $i(x)$ to $1$.