How to Determine the Homology Group of the n-Torus?

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Here's this week's problem!

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Let $\Bbb T^n$ be the $n$-torus, i.e., the product of $n$ circles. Show that the $k^{\text{th}}$ homology group $H_k(\Bbb T^n ; \Bbb Z)$ is isomorphic to $\Bbb Z^{\binom{n}{k}}$ for all $k \le n$.
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No one solved this week's problem. You can find my solution below.

Spoiler

Since $S^1$ has a cellular structure of 1 0-cell and 1 1-cell, $T^n$ has a cellular structure of $\binom{n}{k}$ cells in dimension $k$, for all $k \le n$ (the $k$-cells are of the form $e_{i_1} \times \cdots \times e_{i_k}$, where for each $j$, $e_{i_j}$ is a cell of $S^1$). The cellular boundary is $0$ in every dimension, so $H_k(T^n;\Bbb Z) \approx \Bbb Z^{\binom{n}{k}}$ for all $k \le n$.