Showing that the infinite dimensional sphere $S^{\infty}$ is contractible is rather easy by constructing an explicit contraction (Hatcher gives a nice one). I thought it might be a nice exercise to try and show this using loop spaces and the fact that $\pi_1(S^{\infty})$ is trivial.
Let the loopspace $\Omega X$ be the space of base pointed loops in $X$. For now, assume that the loopspace $ΩS^{\infty}$ is homotopy equivalent to $S^{\infty}$. Then, $\pi_2(S^{\infty})=\pi_1(\Omega S^{\infty})=\pi_1(S^{\infty})=0$. Continuing by induction, we get that every homotopy group of $S^{\infty}$ is trivial and so $S^{\infty}$ is contractible because $S^{\infty}$ is a CW-complex (by Whitehead). As far as I can tell, this argument has no holes in it.
My problem now is in proving the assumption that $\Omega S$ is homotopy equivlanet to $S$. I haven't had too much experience with loopspaces or function spaces in general so I'm struggling somewhat.
One approach might be to find a fibration $F\rightarrow\Omega S^{\infty}\rightarrow S^{\infty}$ and then playing about with long exact sequences in homotopy, but I'm not sure if such a fibration exists. Maybe something like 'If $f\colon [0,1]\rightarrow S^{\infty}$ is a loop, let $p\colon \Omega S^{\infty}\rightarrow S^{\infty}$ be given by $p(f)=f(1/2)$' would work? Is p a fibration? And if so, what do the fibers look like? The best case scenario would be for some fibration to exist with contractible fiber. However, I'm worried that showing such a fiber is contractible would amount to showing that $S^{\infty}$ is contractible, which defeats the object of the exercise.
Best Answer
You might be able to get something using the Freudenthal theorem (though this is a bit overkill admittedly), since it studies the connectedness of the natural map to $\Omega \circ S$. Supposing the assertion that the suspension of $S^\infty$ is again $S^\infty$ (we have to use the limiting behavior of this space somewhere), then if $S^\infty$ is $n$-connected, the map
$\pi_k(S^\infty) \to \pi_k(\Omega S^\infty) = \pi_{k+1}(S^\infty)$
is an isomorphism for $k \le 2n$. Then this is certainly true for $k = n$, which gives that $\pi_{n+1}(S^\infty)$ is trivial as well, and so $S^\infty$ is at least $(n+1)$-connected -- we can continue on to hit all the homotopy groups.