Since the work of Serre in the early 50's on homotopy groups of spheres, it is known that the homotopy group $\pi\_k(S^n)$ is finite, except when $k=n$ (in which case the group is $\mathbb{Z}$), or when $n$ is even and $k=2n-1$ (in which case the group is the direct sum of $\mathbb{Z}$ and a finite group). As a consequence, the stable homotopy groups $\pi\_k^s$ are finite groups for $k>0$, and $\pi\_0^s \cong \mathbb{Z}$. 


The work of Serre was done before anyone knew about stable homotopy theory and chromatic methods, and this makes me wonder about the following questions. 


Question 1: Is it possible to use methods from stable/chromatic homotopy theory to prove finiteness of stable homotopy groups of spheres directly, without having to compute any unstable homotopy groups of spheres? 


Question 2: Is there any philosophical or conceptual reason for why these groups should be finite?



Finiteness of stable homotopy groups of spheres
I agree with Ryan that Serre's proof can be viewed as perfectly conceptual, but here is a modern version. Accept from Serre that the homotopy groups of spheres are finitely generated. Let 
$k\colon S^n \longrightarrow K(\mathbf{Z},n)$ be the canonical map. We know how to rationalize
spaces and maps. The rationalization of $k$ is a map $k\_{0}\colon S^n\_{0}\longrightarrow K(\mathbf{Q},n)$. If $n$ is odd, $k\_0$ is an isomorphism on rational cohomology and therefore
an equivalence. If $n$ is even, a very little cohomological calculation shows that the fiber of $k\_{0}$ is $K(\mathbf{Q},2n-1)$. Since the homotopy groups of spheres are finitely generated,
the kernel of the map on homotopy groups induced by the rationalization $S^n\longrightarrow S^n\_{0}$
is finite in each degree. The rank of the free part is immediate from what I've said about $S^n\_{0}$. Serre's theorem follows. This uses no calculation of unstable homotopy groups except maybe deep down that $\pi\_n(S^n) = \mathbf{Z}$. 


