Regard $S^n$ as a subspace of $\Bbb R^{n+1}$, and consider the reflection $f:S^n\to S^n$, $(x_1,\dots,x_{n+1})\mapsto (-x_1,x_2,\dots,x_{n+1})$. This defines an element of the group $\pi_n(S^n)$. How can we show that the class $[f]\in \pi_n(S^n)$ equals $-[\text{id}_{S^n}]\in \pi_n(S^n)$?
Actually here is one proof: the degree of a reflection is $-1$. And it is known that the degree map $\deg:\pi_n(S^n)\to \Bbb Z$ is an isomorphism, so we get our result.
However I don't want to use the fact that $\deg:\pi_n(S^n)\to \Bbb Z$ is an isomorphism, because this fact is proved using a huge theorem; the Freudenthal suspension theorem.
First, we will give a geometric informal argument showing that the reflection and identity are inverses in $\pi_n(S^n)$. This shall serve as intuition for when we do the formal argument. The class $[f]+[\text{id}_{S^n}]\in\pi_n(S^n)$ is represented by the map $$ S^n \overset{\text{pinch map}}{\longrightarrow} S^n \vee S^n \overset{f\vee \text{id}_{S^n}}{\longrightarrow}S^n.$$ Thus, $[f]+[\text{id}_{S^n}]$ is represented by a map that quotients out the equator and maps the lower hemisphere $S^n$ via the quotient map and wraps the upper hemisphere around $S^n$ via the reflection. Since the upper and lower hemisphere wraps around with different orientations, we can "cancel them out" by unwrapping yielding the constant map at the basepoint.
Let us now do the argument more formally. We will have to use a specific model for the higher homotopy group. $\pi_n(X,x_0)$ can be described as the set of homotopy classes of maps $([-1,1]^n,\partial [-1,1]^n)\to (X,x_0) $ with the opreation given by $$ (f+g)(t_1,\ldots,t_n)= \begin{cases} f(2t_1+1,t_2,\ldots,t_n) & -1 \leq t_1 \leq 1 \\ g(2t_1-1,t_2,\ldots,t_n) & 0 \leq t_1 \leq 1 \end{cases} $$ This is the definition that Hatcher uses slightly modified using the obvious homeomorphism $[-1,1]\cong[0,1]$. We have that $-g(t_1,\ldots,t_n)=g(-t_1,\ldots,t_n)$. This is shown similarly as for the fundamental group. Futhermore, this corresponds to the unwrapping in our informal geometric description.
The problem with this definition in our case is that we consider elements of $\pi_n(S^n)$ as homotopy classes of maps out of $(S^n,*)$, where $S^n$ is considered a subspace of $\mathbb{R}^{n+1}$, and not out of $([-1,1]^n,\partial [-1,1]^n$). So we will have to translate them using a homeomorphism $\psi: [-1,1]^n/\partial [-1,1]^n\to S^n$. Combining this and this, we get the homeomorphism: $$ \psi:t \mapsto \left(2\frac{\vert\vert t\vert \vert_\infty}{\vert\vert t\vert \vert_2} \sqrt{1-(\vert\vert t\vert \vert_\infty)^2}t,2(\vert\vert t\vert \vert_\infty)^2-1\right) $$ Then a based map $(S^n,*)\to(S^n,*)$ is considered as an element of $\pi_n(S^n)$ by precomposition with $\psi$.
Let $\tilde{f}:([-1,1]^n,\partial [-1,1]^n) \to ([-1,1]^n,\partial [-1,1]^n)$ be given by $\tilde{f}(t_1,\ldots,t_n)=(-t_1,t_2,\ldots,t_n)$. Check that $\psi\circ \tilde{f}=f\circ \psi$. From this, it follows that $[f\circ \psi]$ and $[\psi]$ are inverses. We conclude that $[f]$ and $[\text{id}_{S^n}]$ are inverses.