I am trying to fill in some details of an argument. It might be well-known but I don't know any written references. I think I heard it from a lecture by John Morgan a few years ago and we trying to remember his argument.
Claim: $\pi_{n+1}(S^n) = \mathbb{Z}_2$ for $n \geq 3$.
Observe that if $n=2$, $\pi_3(S^2) = \mathbb{Z}$.
Sketch of proof: Let $f:S^{n+1} \to S^n$; we may assume that this is a smooth map as continuous maps are homotopic to smooth ones in this setting. In that case, let $f$ be transverse to some point, say, the basepoint $p \in S^n$. Then $f^{-1}(p)$ is a link $L_f \subset S^{n+1}$. Due to the high dimensions, the link can be isotoped to be unlinked and unknotted. Note that $p$ is a 0-manifold in $S^n$ and it has a trivial normal bundle which we may pull back to $L_f$; so $L_f$ has a framing.
Lemma: If a framed circle is cobordant to two other framed circle$, then it is framed cobordant to the connected sum of the two framed circles.
Evidently, they are cobordant and I think it isn't much trouble to show framed cobordant as well. With this established, then we can reduce our study from framed links to framed knots.
Lemma: $f,g:S^{n+1} \to S^n$ are homotopic if and only if $L_f$ and $L_g$ are framed cobordant.
$\Longrightarrow$ Let $H:S^{n+1} \times I \to S^n$ be a homotopy; we may assume that $f,g,H$ are all transverse to the basepoint. This is by Sard's theorem which tells us that the regular values of smooth maps are open and dense. So we can find some point which all three maps are transverse to. Then the homotopy $H$ constructs a framed cobordism for us inside of $S^{n+1} \times I$ between $L_f$ and $L_g$.
$\Longleftarrow$ Here is where I am most unsure. Suppose $W$ is a framed cobordism for $L_f$ and $L_g$. I imagine we could embed this into $S^{n+1} \times I$ with $L_f \subset S^{n+1} \times \{0\}$ and $L_g \subset S^{n+1} \times \{1\}$ but I'm not sure how that would give us a homotopy. It seems one would want to let the cobordism "guide" the construction of a homotopy. For instance, $H_t^{-1}(p) = W \cap (S^{n+1} \times \{t\})$. But this only would help us define a map on a link; to extend it to $S^{n+1}$ seems to require much more. Supposing we have defined $H$ in a neighborhood of $W$ and $H$ is transverse to $p$. Then we could maybe move $p$ around among the regular values and extend $H$ in a similar fashion as before.
Assuming we have this lemma, then the rest is easy. The first lemma reduces the question of computing $\pi_{n+1}(S^n)$ to considering framed circles. A frame at a point is a linearly independent basis of $n$ vectors; we can identify this with an element of $GL^+(n,\mathbb{R})$ which deformation retracts to $SO(n)$. So we just need to know that $\pi_1(SO(n))=\mathbb{Z}_2$ which can be proven with some straight forward homotopy theory.
As a sidenote, I was interested in this because the prime $p=2$ seems to always be the most troublesome prime and among the homotopy groups, 2-torsion immediately appears. By that, I mean $\pi_k(S^n)=0$ for $k < n$ and $\pi_n(S^n) = n$. And then $\pi_{n+1}(S^n) = \mathbb{Z}_2$ for $n \geq 3$.