A modified version of your 1: If you have the Hurewicz theorem for $\pi_n(S^n)$, then you also have it for $k < n$, so you don't need the geometric argument for $\pi_k(S^n)=0$ when $k < n$.

Alternatively, after arguing geometrically that $\pi_k(S^n)=0$ when $k < n$, you can use much the same idea to show that any map $S^n\to S^n$ is homotopic to one that takes everything to the basepoint except some little disks that are mapped in a standard way. From there you can go to seeing that $\pi_n(S^n)$ is generated by the class of the identity map. And then you don't need the full strength of Hurewicz, but just the fact that all the multiples of the identity have different effects on $H_n(S^n)$. (The first part of this is pretty close in spirit to the framed cobordism argument indicated by Chris Gerig.)

A modified version of your 1: If you have the Hurewicz theorem for $\pi_n(S^n)$, then you also have it for $k

Alternatively, after arguing geometrically that $\pi_k(S^n)=0$ when $k