Proof:

Let $F: I \times I \longrightarrow S^1$ be the homotopy between $\gamma_1$ and $\gamma_2$. Since $I \times I$ is star shaped with respect to $(0,0)$ and $F(0,0)= 1 =$   ex$(0),$ the lifting lemma gives a unique lift ${\tilde F} :\: I \times I \longrightarrow \mathbb{R}$ with ${\tilde F}(0,0) = 0$. The image $F(s,0)$ is a connected subset of $\mathbb{R}$ as $s$ runs from 0 to $1$ and exp${\tilde{F}(s,0)} = F(s,0) = 1$ for all $s\in [0,1]$. So $\tilde{F}(s,0)$ is integer valued and hence constant. From $\tilde{F}(0,0) = 0$ we conclude that $\tilde{F}(s,0)=0$ for all $s\in [0,1]$. In particular the lifts ${\tilde \gamma}_1$ and ${\tilde \gamma}_2$ both start at the origin and so

   deg$$\;\gamma_1 = {\tilde \gamma}_1(1),$$   deg$$\;\gamma_2 = {\tilde \gamma}_2(1).$$

Our job will be over if we show that ${\tilde \gamma}_1(1) = {\tilde \gamma}_2(1)$. Well, ${\tilde F}$ must map the connected set $\{(s,1)/ 0 \leq s \leq 1 \}$ onto a connected subset $J$ of $\mathbb{R}$ and since

   ex$${\tilde F}(s,1) = F(s,1) = 1,$$

this connected subset $J$ must be a subset of $\mathbb{Z}$ and hence reduces to a singleton which means

$${\tilde F}(s,1) = {\tilde F}(0,1),$$    for all $$s \in [0,1]$$

Setting $s = 0$ and $1$ we see that

$${\tilde \gamma}_2(1) = {\tilde F}(1,1) = {\tilde F}(0,1) = {\tilde \gamma}_1(1),$$

thereby completing the proof that the map $\phi:[\gamma] \mapsto$   deg${\gamma}$ is well defined.