I'm supposed to show that for the Torus $T^d = (S^1)^d$ we have the formula:
$$ H_n(T^d) = \bigoplus_{i = 0}^d \bigoplus_{j = 1}^\binom{d}{i} H_{n-i}(\text{point}) = \begin{cases} R^\binom{d}{n} &0\leq n \leq d \\ 0 & \text{else} \end{cases} $$
where $H$ is any ordinary homology theory. We have already proofed and its probably a consequence out of that the following things:
If $X$ is locally compact then $H_n(S^d \times X) = H_n (X) \times H_{n - d}(X)$ and we know the homology of the sphere i.e $H_n(S^d) = \begin{cases} R & d \neq 0,\ n = 0, d \\ 0 & d \neq 0,\ n \neq 0, d \\ R \oplus R & d = 0,\ n = 0 \\ 0 & d = 0,\ n \neq 0 \end{cases} $
How can I prove the homology of the torus out of that?
This is a straightforward proof by induction. For the base case $d = 1$ we directly see that $H_0(S_1) \cong H_1(S^1) \cong R = R^\binom{1}{0} = R^\binom{1}{1}$. Adopting the convention that $\binom{a}{b} = 0$ if $b < a$ or $b > a$ and assuming now that we've proven the identity up to some fixed $d - 1 \geq 1$, we have for all $n$ that \begin{align} H_n(T^d) &= H_n(S^1 \times T^{d - 1}) \\ &\cong H_n(T^{d - 1}) \oplus H_{n - 1}(T^{d - 1}) \\ &\cong R^\binom{d - 1}{n} \oplus R^\binom{d - 1}{n - 1} \\ &\cong R^\binom{d}{n} \end{align} where we use the binomial identity $\binom{d - 1}{n} + \binom{d - 1}{n - 1} = \binom{d}{n}$ for the last step.