A question about the Homotopy groups of Disks

Question

It is not hard to see that $\pi_1(D^2) = 0$, but is it sound to say that $\pi_n(D^{n+1}) =0$ for all $n \in \mathbb{N}$?

Answer 1

Any disk $\Bbb D^n=\{x \in \Bbb R^n\mid\Vert x\Vert \leq 1\}$ is contractible, ie. there is a homotopy equivalence to the one-point space $\{*\}$: We have that $\{0\}\subseteq \Bbb D^n$ is a deformation retract by means of the homotopy $h:\Bbb D^n \times [0,1]\rightarrow \Bbb D^n, (x,t) \mapsto tx$.

But then each homotopy group of $\Bbb D^n$ vanishes, since $$\pi_k(\Bbb D^n) = [\Bbb S^k, \Bbb D^n] \cong [\Bbb S^k, \{\ast\}] = \pi_k(\{\ast\}) = 0$$