Note: Since the degree of a map is independent of the base-point I'll speak loosely and just say $\pi_1(S^1)$.
One definition of the degree of a map $f$ from the circle to itself is the number $k$ determined by where $f_\ast$ sends the generator $1$, since $\mathbb{Z}$ is isomorphic to $\pi_1(S^1)$.
i.e. $k$ in $f_\ast(1) = k\cdot1$.
From this definition I'm interested in proving that for two maps $f,g: S^1\rightarrow S^1$, $\deg(f\circ g) = \deg(f)\cdot \deg(g)$.
Denote $\deg(f) = m$ and $\deg(g) = n$.
So then I know $(f\circ g)_\ast = f_\ast \circ g_\ast$.
Then $$(f\circ g)_\ast (1) = f_\ast \circ g_\ast(1)$$ $$= f_\ast(n\cdot1)$$ $$= m\cdot n \cdot 1$$ $$ = mn$$
Therefore $\deg(f\circ g) = \deg(f)\cdot \deg(g)$.
Is this proof valid?
Let $f$ be a continuous map from $S^1\longrightarrow S^1$. It induces a map on $\pi_1$; that is a map $f_*: \pi_1(S^1)\cong \mathbb Z \rightarrow \pi_1(S^1)\cong \mathbb Z $. Now the only group homomorphism from $\mathbb Z$ to $\mathbb Z$ is multiplication by an integer $k_f$. This $k_f$ is the degree of $f$.
The fact that $(f\circ g)_*=f_*\circ g_*$ is a functorial property of the functor $\pi_1$. Hence $\deg(f\circ g)$ is just multiplication by the integer $k_g$ then by the integer $k_f$.