How does the Eckmann-Hilton argument show that higher homotopy groups are commutative?
I can easily follow the proof on Wikipedia, but I have no good mental picture of the higher homotopy groups, and I can't see how to apply it. Wikipedia mentions this application in one sentence, with no explanation.
(Motivation: I'm thinking of giving a talk on algebraic topology for a general audience of math majors. The purpose would be to try to explain how purely "algebraic" methods can be used to gain serious insight into purely "geometric" problems. This would be a great example, if I could do the proof!)
Thanks in advance!

Recall that the Eckmann-Hilton argument says that if you have two different unital monoid structures on a set such that both are homomorphisms for the other and both have the same unit, then they're equal and commutative. We want to apply that to see that higher homotopy groups are commutative.
The two different monoid structures we want to consider are 'vertical' and 'horizontal' concatenation of maps from $I^n$: if I have two maps $f,g:I^n\to X$ for some topological space $X$ so that $f(\partial I^n)=g(\partial I^n)=x_0$, then I can concatenate them in the following ways as long as $n>1$:
$f\cdot g=\begin{cases} f(2x,y,\cdots) &\text{ for } 0\leq x\leq 1/2\\ g(2x-1,y,\cdots) &\text{ for } 1/2\leq x\leq 1 \end{cases}$
$f \star g= \begin{cases} f(x,2y,\cdots) &\text{ for } 0\leq y\leq 1/2\\ g(x,2y-1,\cdots) &\text{ for } 1/2\leq y\leq 1 \end{cases}$
Now, if I can show that these are homomorphisms for each other, then I'll have that $\pi_n$ is commutative for $n>1$. But this is easy: both $(f\cdot g)\star (h\cdot i)$ and $(f\star h)\cdot (g\star i)$ are the following map:
$(x,y,...)\mapsto\begin{cases} f(2x,2y,\cdots) &\text{ for }(x,y)\in [0,1/2]\times [0,1/2]\\ g(2x-1,2y,\cdots) &\text{ for } (x,y)\in [1/2,1]\times [0,1/2]\\ h(2x,2y-1,\cdots) &\text{ for } (x,y)\in [0,1/2]\times [1/2,1]\\ i(2x-1,2y-1,\cdots) &\text{ for } (x,y)\in [1/2,1]\times [1/2,1] \end{cases}$
Now, by the Eckmann-Hilton argument, these two sorts of composition are the same and are in fact commutative.