If $f:S^1 \to S^1$ is continuous and deg$(f)=\pm 1$, then $f$ is a homotopy equivalence. Let $f:S^1 \to S^1$ be a continuous function such that its degree is deg$(f)=\pm 1$. Using Whitehead theorem, I need to prove that $f$ is a homotopy equivalence. 
My Attempt:
What I tried to do is to use the corollary mentioned in Wikipedia, which is obtained by combining Whitehead theorem with Hurewicz theorem, and has the following statement: "a continuous map $f : X \to Y$ between simply connected CW complexes that induces an isomorphism on all integral homology groups is a homotopy equivalence."
Since deg$(f)=\pm 1$, the function $f_*:\widetilde{H}_n(S^1) \to \widetilde{H}_n(S^1)$ is multiplication by $\pm 1$, so that $f$ is the identity or the antipodal map. In both cases, $f$ would be an homeomorphism, and then $f_*$ would be an isomoprhism for all $n$, and by the corollary mentioned above, $f$ is a homotopy equivalence.
I would like to know if my proof is correct or how can I prove this correctly. Can this be proved directly from Withehead theorem?
Diagram added:
$\require{AMScd}$
\begin{CD}
H_n(S^n) @>{f_*}>> H_n(S^n) \\
@VVV{h_*} @VVV{h_*}\\
\pi_n(S^n) @>{f_*}>> \pi_n(S^n)
\end{CD}

The circle $S^1$ is not simply connected, so you cannot use the corollary. Instead you need to use Whitehead's Theorem which states that a continuous map between CW complexes is a homotopy equivalence if and only if it induces isomorphisms on all homotopy groups. 
As $S^1$ is a CW complex and $\pi_n(S^1) = 0$ for $n \geq 2$, we see by Whitehead's theorem that $f : S^1 \to S^1$ is a homotopy equivalence if and only if the induced map $f_* : \pi_1(S^1) \to \pi_1(S^1)$ is an isomorphism. 
Note that $\pi_1(S^1) \cong \mathbb{Z}$ and the map $f_*$ is nothing but multiplication by $\deg f$. So $f_*$ is an isomorphism, and hence $f$ is a homotopy equivalence, if and only if $\deg f = \pm 1$.
