Expressing the Hilbert polynomial of a complete intersection as the difference of two Hilbert polynomials Suppose $ Z_{i} = \text{Proj}[x_{0},\dots,x_{n}]/(F_{i}) $ is a hypersurface of degree $ d_{i} $  for $i=1,2,\cdots,r$ in projective space over a field $ k. $
Consider $ \Gamma = \text{Proj}[x_{0},\dots,x_{n}]/(F_{1},\dots,F_{r}), $ and let us denote the Hilbert polynomial of such a complete intersection by $ 
P_{d_{1},\dots,d_{r}}(v). $ 
For any subset $$ I = \lbrace i_{1},\dots, i_{k} \rbrace \subset \lbrace 1,2,\dots,r \rbrace, $$
Now suppose that $$ P_{d_{1},\dots,d_{r}}(v) =  \sum_{I \subset \lbrace 1,\dots,r \rbrace} (-1)^{|I|} \binom{n-v-d_{I}}{n} = \sum_{k=0}^{r} (-1)^{k}\sum_{|I|=k} \binom{n+v-d_{r}-d_{I}}{n}, $$ where the sum ranges over all subsets of $ \lbrace 1,2,\dots,r \rbrace, $ $ d_{I} = \sum_{\alpha}^{k} d_{i_{\alpha}}, $ and $ |I| $ denotes the number of elements in $ I. $
Then it is claimed that $$ P_{d_{1},\dots,d_{r}}(v) = P_{d_{1},\dots,d_{r-1}}(v) - P_{d_{1},\dots,d_{r-1}}(v-d_{r}). $$
I have not been able to establish the truth of this, and attempting to show compute $$ P_{d_{1},\dots,d_{r-1}}(v) - P_{d_{1},\dots,d_{r}}(v), $$ by brute force yields $$ \sum_{|I|=r}(-1)^{r+1} \binom{n+v-d_{I}}{n} $$ I think that this is not equal to $ P_{d_{1},\dots,d_{r-1}}(v-d_{r}). $
 A: Given the definition $P_{d_1,\ldots, d_r}(v) = \sum_{I \subseteq \{1,\ldots, r\}} (-1)^{|I|} \binom{n+v-d_I}{n}$, 
\begin{align} P_{d_1,\ldots, d_r}(v) - P_{d_1,\ldots, d_{r-1}}(v) &= \sum_{d_r \in I} (-1)^{I}\binom{n+v-d_I}{n} \\&= \sum_{J \subseteq \{1,\ldots, r-1\}} (-1)^{|J|+1} \binom{n + v - d_J - d_r}{n} = - P_{d_1,\ldots, d_{r-1}}(v-d_r).\end{align}
Presumably, this step is used in an inductive proof to calculate the Hilbert polynomial of $\Gamma$. A more conceptual way of understanding this formula is via the Koszul complex: if $S = k[x_0,\ldots, x_n]$ is the coordinate ring of projective space, then for $(F_1,\ldots, F_r)$ a complete intersection where $F_i$ has degree $d_i$, let $E = \oplus_{i=1}^r S(-d_i)$. Then we have an exact sequence
$$ 0 \leftarrow S/(F_1,\ldots, F_r) \leftarrow S \leftarrow E \leftarrow \wedge^2 E \leftarrow \wedge^3 E \leftarrow \cdots\leftarrow \wedge^r E \leftarrow 0$$
where the wedge power is over $S$. 
Concretely, as an $S$-module $\wedge^k E = \sum_{|I| = k} S(-d_I)$, which gives your formula.
