Recall that the exponentiated operator may be expanded as
Inserting this into the energy equation (3.7) we obtain
which becomes, after distributing terms
Note that 
is at
most a two-particle operator and that 
is at least a one-particle
excitation operator. Then, assuming that the reference wavefunction is a
Slater determinant constructed from a single-particle basis, Slater's
rules state that matrix elements of the Hamiltonian
between determinants that differ by more than two single-particle states
(i.e. spin orbitals) are zero. Thus, the fourth term on the LHS of the
above equation contains, at the least, three-fold excitations. Thus, that
matrix element (and all higher-order elements) necessarily vanish.
The energy equation then becomes
This is the natural truncation of the CC energy equation (an analogous
phenomenon occurs for the amplitude equations (3.8)). This
truncation depends only on the form of and not on that of
or on the number of electrons, and, therefore, is exact even if
is truncated to a particular excitation level.