I am looking for a hint to prove the following identity. Let $A\in \mathbb{R}^{d\times n}$ ($n\geq d$) and $A_{-i}$ be $A$ without the $i$th column $a_i$. Assume both matrices are full-rank. Then show that $$\frac{\text{det}\left(A_{-i}A_{-i}^T\right)}{\text{det}(AA^T)} = 1-a_{i}^T(AA^T)^{-1}a_i. $$
In particular, Sylvester's identity might be relevant, but it's not obvious to me.
Viewing the rhs as a $1\times 1$ matrix, Sylvester's identity lets us rewrite the problem as
$$ \frac{\text{det}\left(A_{-i}A_{-i}^T\right)}{\text{det}(AA^T)} = 1-a_{i}^T(AA^T)^{-1}a_i=\det(1-a_{i}^T(AA^T)^{-1}a_i)=\det(I_d - (AA^T)^{-1}a_ia_i^T), $$ and multiplying through by $\det(AA^T)$, $$ \det(A_{-i}A^T_{-i})=\det(AA^T)\det(I_d - (AA^T)^{-1}a_ia_i^T)=\det(AA^T-a_ia_i^T). $$ In this form the identity follows by writing $AA^T$ as $\sum_j a_ja_j^T$.