Proof:

Suppose $\gamma_1$ and $\gamma_2$ are two loops at $1$ with lifts ${\tilde \gamma}_1, {\tilde\gamma}_2$ starting at origin. Then the path $\tilde{\gamma}$ given by $\tilde{\gamma}(t) = \tilde{\gamma_1}(t) + \tilde{\gamma_2}(t)$ also starts at the origin and satisfies

   ex$$\;{\tilde \gamma}(t) =$$   ex$$\;{\tilde \gamma_1}(t) \cdot$$   ex$$\;{\tilde\gamma_2}(t) = \gamma_1(t) \cdot \gamma_2(t).$$

Hence $\tilde{\gamma}$ is the unique lift of $\gamma_1(t)\cdot\gamma_2(t)$ whereby,

   deg$$(\gamma_1(t) \gamma_2(t)) = {\tilde\gamma}(1) = {\tilde\gamma_1}(1) + {\tilde\gamma_2}(1) =$$   deg$$\;{\gamma_1} +$$   deg$$\;{\gamma_2}.$$

Thus $\phi( [\gamma_1 \cdot \gamma_2]) = \phi([\gamma_1]) + \phi ([\gamma_2])$. From corollary (12.2), $[\gamma_1 \cdot \gamma_2] = [ \gamma_1 \ast \gamma_2]= [\gamma_1][\gamma_2]$ whence $\phi([\gamma_1][\gamma_2]) = \phi([\gamma_1])+ \phi( [ \gamma_2])$ which means that $\phi$ is a group homomorphism.

Surjectivity of $\phi$ is easy to see. Let $n \in \mathbb{Z}$ be arbitrary and ${\tilde \gamma}(t) = nt.$ Then ${\tilde\gamma}$ is the unique lift of $\gamma(t) =$   ex${\tilde\gamma}(t)$ starting at the origin so that $\phi([\gamma]) = {\tilde\gamma}(1) = n$. We now show that the group homomorphism $\phi$ is injective. Suppose $\gamma_1,\gamma_2$ are two loops at $1$ in $S^1$ such that deg $\;\gamma_1$ = deg $\;\gamma_2$. Then ${\tilde\gamma_1}(1) = {\tilde \gamma_2}(1)$, where ${\tilde\gamma_1}$ and ${\tilde\gamma_2}$ are the lifts of $\gamma_1$ and $\gamma_2$ starting at the origin. Since $\mathbb{R}$ is convex and the two curves $\tilde{\gamma}_1$ and $\tilde{\gamma}_2$ have common end points, they are homotopic. That is to say, there exists a continuous function ${\tilde F} : I \times I \longrightarrow \mathbb{R}$ such that

$${\tilde F}(0,t) = \tilde{\gamma_1}(t),\quad \tilde{F}(1,t) = \tilde{\gamma_2}(t); t \in [0, 1]$$

$$\tilde{F}(s,0) = 0, \quad\tilde{F}(s,1) = \tilde{\gamma_1}(1) = \tilde{\gamma_2}(1), s \in [0, 1].$$

The function $F:[0, 1]\times [0, 1]$ given by

$$F(s,t)=$$   ex$$\tilde{F}(s,t)$$

is then a homotopy between $\gamma_1$ and $\gamma_2$ and we have shown that $\deg \gamma_1= \deg \gamma_2$ implies $[\gamma_1]= [ \gamma_2]$. This suffices for a proof.