Let us give a simple proof of Thouless' theorem. The theorem states that we can make a linear combination av particle-hole excitations with respect to a given reference state \( \vert c\rangle \). With this linear combination, we can make a new Slater determinant \( \vert c'\rangle \) which is not orthogonal to \( \vert c\rangle \), that is $$ \langle c|c'\rangle \ne 0. $$ To show this we need some intermediate steps. The exponential product of two operators \( \exp{\hat{A}}\times\exp{\hat{B}} \) is equal to \( \exp{(\hat{A}+\hat{B})} \) only if the two operators commute, that is $$ [\hat{A},\hat{B}] = 0. $$