How to prove $\deg( f\circ g) = \deg(f) \deg(g) $? If $ f,g:S^1 \rightarrow S^1$ continuous maps then 
\begin{equation*}
\deg( f\circ g)= \deg(f)\deg(g).
\end{equation*}
Unfortunately, i haven't made any progress in solving it. I've tried considering the lift of f,g and  $f\circ g$ but i can't see how to continue.
I would appreciate any hint you have in mind.

From OP's comments: the degree of $f : S^1 \to S^1$ is defined as $\tilde{f}(1) - \tilde{f}(0)$, where $\tilde{f} : [0,1] \to \mathbb{R}$ makes the following diagram commute, where the vertical arrows are projections $t \mapsto e^{2i\pi t}$:
$$\require{AMScd}
\begin{CD}
[0,1] @>{\tilde{f}}>> \mathbb{R} \\
@VVV @VVV \\
S^1 @>f>> S^1
\end{CD}$$
(this does not depend on the choice of $\tilde{f}$).
 A: If you are using the "lifting definition", just consider this commutative diagram:
$$
\begin{array}{ccccc}
\Bbb R & \stackrel{G}{\longrightarrow} & \Bbb R & \stackrel{F}{\longrightarrow} & \Bbb R \\
\downarrow & & \downarrow & & \downarrow \\
S^1 &\stackrel{g}{\longrightarrow} &S^1& \stackrel{f}{\longrightarrow} & S^1  
\end{array}.
$$
The map $G$ sends an integer $k$ to $\text{deg}(g) \times k$ (it sends $1$ to $\text{deg}(g)$, $2$ to $2\times \text{deg}(g)$ and so on) and $F$ to $\text{deg}(f)\times k$. So, the composition $F\circ G$ sends $1$ to $\text{deg}(g)\times \text{deg}(f)$.
A: First note:
1) that two continuous maps $f,g:S^1 \rightarrow S^1$ have the same degree iff they are homotopic. 
2) If $\deg(f)=n$ and $\deg(g)=m$, then $f \simeq z^n$ and $g \simeq z^m$
3) Composition of maps behaves well with respect to composition, i.e $(f \circ g)\simeq (z^n \circ z^m)$
Now suppose $\deg(f)=n$ and $\deg(g)=m$. We use 1) 2) and 3) to obtain $\deg(f \circ g)=\deg(z^n \circ z^m)$
I think this can be an easy way to answer your question without using advanced machinery.
Remark on notation: by $z^n$ I mean $(cos(t),sin(t) \in S^1 \rightarrow (cos(nt),sin(nt)) \in S^1$ ( walking around n times in the image for each one in the domain).
