If the operators do not commute, we need to resort to the Baker-Campbell-Hauersdorf. This relation states that $$ \exp{\hat{C}}=\exp{\hat{A}}\exp{\hat{B}}, $$ with $$ \hat{C}=\hat{A}+\hat{B}+\frac{1}{2}[\hat{A},\hat{B}]+\frac{1}{12}[[\hat{A},\hat{B}],\hat{B}]-\frac{1}{12}[[\hat{A},\hat{B}],\hat{A}]+\dots $$ From these relations, we note that in our expression for \( |c'\rangle \) we have commutators of the type $$ [a_{a}^{\dagger}a_{i},a_{b}^{\dagger}a_{j}], $$ and it is easy to convince oneself that these commutators, or higher powers thereof, are all zero. This means that we can write out our new representation of a Slater determinant as $$ |c'\rangle=exp\left\{\sum_{a>F}\sum_{i\le F}C_{ai}a_{a}^{\dagger}a_{i}\right\}| c\rangle=\prod_{i}\left\{1+\sum_{a>F}C_{ai}a_{a}^{\dagger}a_{i}+\left(\sum_{a>F}C_{ai}a_{a}^{\dagger}a_{i}\right)^2+\dots\right\}| c\rangle $$