= Solution
Use <singular homology> with <integer> coefficients. The <Künneth theorem> for triangulable spaces gives a split short exact <sequence> with tensor terms $\bigoplus_{i+j=k}H_i(X)\otimes H_j(Y)$ and torsion terms $\bigoplus_{i+j=k-1}\operatorname{Tor}_1(H_i(X),H_j(Y))$. When one factor's <homology groups> are free abelian, the torsion terms vanish and the tensor map is an <isomorphism>.
The <circle> has $H_0(S^1)=H_1(S^1)=\mathbb Z$ and all other <groups> zero. Inductively, the <homology groups> of a <torus> are free, and
$$
H_k(T^{n-1}\times S^1)\cong H_k(T^{n-1})\oplus H_{k-1}(T^{n-1}).
$$
Pascal's identity gives
$$
\boxed{H_k(T^n;\mathbb Z)\cong\mathbb Z^{\binom nk}\quad(0\leq k\leq n),\qquad H_k=0\quad(k>n).}
$$
Geometrically the generators correspond to choosing $k$ of the $n$ <circle> factors and taking their product, consistent with the cellular model having one $k$-cell for each such <subset> and cancelling opposite boundary faces.
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