For a monomial ideal $I$ of a polynomial ring $S$ with degree $b$, Theorem 1.34 of the ``Combinatorial Commutative Algebra" says that
$$\beta_{i,b}(I) = \beta_{i+1,b}(S/I)$$
with a proof stating that
use a minimal free resolution of $I$ achieved by snipping off the copy of $S$ occurring in homological degree 0 of a minimal free resolution of $S/I$.
This is immediate if the homological degree 0 part of the free resolution, say $F_{0}$, is $S$; however, how this proof is applied when $F_{0}=\bigoplus_{i=1}^{n}S$ with $n>1$? It seems that $\text{im}(F_{1}) \subset F_{0}$ may be a submodule of $F_{0}$ which is not equivalent to $I$. Could you give me any hint to understand this proof?
Perhaps another$^1$ proof helps to illuminate things: consider the short exact sequence
$$0 \longrightarrow I\longrightarrow S \longrightarrow S/I\longrightarrow 0$$
and take the long exact sequence for $\mathrm{Tor}^S(-,k)$. Since $S$ is projective its Tor-groups vanish, and the connecting morphism induces natural isomorphisms
$$\delta : \mathrm{Tor}^S_{j+1}(S/I,k) \longrightarrow \mathrm{Tor}^S_j(I,k).$$
Taking dimension gives your equality, if you recall that the Tor-dimension gives you the dimension of generators of a minimal resolution.
1: One can think of this as another incarnation of "snipping off the copy of $S$", as the SES above is precisely the truncation of the beginning of the minimal resolution of $S/I$.