Is the Galois group of the splitting field for $\prod{(X^{p_i}-1)}$ over $\mathbb{Q}$ always $\mathbb{Z}_{p_1-1} \times \dots \times \mathbb{Z}_{p_k-1}$, where $p_1, \ldots, p_k$ are distinct primes?
I think the splitting field would be $\mathbb{Q}(\zeta_{p_1}, \ldots \zeta_{p_k})$. Since $Gal(\mathbb{Q}(\zeta_{p})/\mathbb{Q}) \cong \mathbb{Z}_{p-1}$ I feel like this would make sense.
Yes. We have $\mathbf{Q}(\zeta_n,\zeta_m)=\mathbf{Q}(\zeta_{\text{lcm}(n,m)})$ (composite of cyclotomic extensions, see here). We get that your splitting field is $\mathbf{Q}(\zeta_{\prod p_i})$. The Galois group is now equal to $(\mathbf{Z}/\prod p_i \mathbf{Z})^\times \cong \prod (\mathbf{Z}/p_i\mathbf{Z})^\times$ by the Chinese remainder theorem.