The results are listed here:
Is there an intuitive way to understand these results?
In particular, why would the higher homotopy group be the trivial group?
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$\begingroup$Let $\mathbb{T}^n$ denote the $n$-torus. This is the product of $n$ circles, i.e., $\mathbb{T}^n=(S^1)^n$. Thus, $\pi_1(\mathbb{T}^n)=\prod^{n}\pi_1(S^1)=\mathbb{Z}^n$. Since $\mathbb{T}^n$ admits a universal cover $\alpha:\mathbb{R}^n\to\mathbb{T}^n$, for $i\geq 2$, the homomorphisms $\pi_i(\alpha):\pi_i(\mathbb{R}^n)\to\pi_i(\mathbb{T}^n)$ are isomorphisms. Since $\mathbb{R}^n$ is homotopy equivalent to the point, $\pi_i(\mathbb{T}^n)$ is the trivial group for $i\geq 2$.
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