$\prod_{i=0}^n x_i^{s_i} = \prod_{i=0}^n (e^{{\frac{2\pi i }{d} \sum_{i=0}^n b_{ij}h_j}} x_i)^{s_i}$. This is equivalent to $\sum_{i,j} s_ib_{ij}h_j$ being a multiple of $d$.

Birational Geometry of BHK Mirrors
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We now give a Theorem of Bini, written in our notation (Theorem 3.1 of [@Bi]):

Let all the notation be as above. Then the hypersurfaces $X_A$ and $X_{A^T}$ are birational to the quotients of the Fermat variety $X_{dI}/ (((\phi_B)_*)^{-1}(J_{F_A})/J_{F_{dI}})$ and $X_{dI}/ (((\phi_{B^T})_*)^{-1}(J_{F_{A^T}})/J_{F_{dI}})$, respectively.

We now give a few comments about the proof of the above theorem. It is proven via composing $\phi_B$ with the map $$\begin{aligned}
q_A : W{\mathbb{P}}^n(q_0, \ldots, q_n) &\dashrightarrow {\mathbb{P}}^{n+1};\\
(x_0:\ldots: x_n) &\longmapsto \left(\prod_j x_j : \prod_j x_j^{a_{1j}}:\ldots:\prod_j x_j^{a_{nj}}\right).
\end{aligned}$$ Note that the composition $q_A \circ \phi_B: X_{dI} \dashrightarrow {\mathbb{P}}^{n+1}$ gives the map $$(y_0:\ldots:y_n) \longmapsto \left(\prod_j y_j^{q_j'}: y_1^d:\ldots: y_n^d\right).$$ Letting $m = \gcd(d, q_1', \ldots, q_n')$, we describe the closure of the image as $$\overline{M_A} := Z\left(\sum_{i=1}^n u_i, u_0^{d/m} = \prod_{i=1}^n u_i^{q_i'/m}\right) \subset {\mathbb{P}}^{n+1}.$$ Bini then proves that the map $q_A\circ\phi_B$ is birational to a quotient map, which in our notation implies the birational equivalence $$\overline{M_A} \simeq X_{dI} / (\phi_B^{-1}(SL(F_A))/ J_{F_{dI}}).$$

Bini then refers the reader to the proof of Theorem 2.6 in [@BvGK] to see why the other two 