Let $k$ be a field and let $X= \mathbb{P}_{k}^{r}$ be the projective space (as a scheme) of dimension $r$ over $k$. Let $\mathcal{O}(d)$ denote the degree $d$ twisted structure sheaf. Then we define the Euler characteristic of any coherent sheaf $\mathcal{F}$ by $$ \chi(\mathcal{F}) = \sum (-1)^{i} \text{dim}_{k} H^{i}(X, \mathcal{F}). $$ I am trying to show that the terms $\chi(\mathcal{F}(d))$ give a polynomial in $d$. I have reduced this to showing that it holds for the case that $\mathcal{F} = \mathcal{O}(d)$. This seems to be the standard approach. However, several sources have all then said something very confusing. They've written that $\chi(\mathcal{O}(d))$ is given by the binomial coefficient $$ {d+r \choose d} $$ I can see why this would be true for sufficiently large $d$, since the only non-vanishing cohomology is in degree $0$. But for arbitrary $d$, we can have cohomology appearing for $d \leq -r-1$. Indeed the binomial coefficient would not even be defined for $d< 0$. But it seems that sources I've read don't qualify this statement by saying that it only holds for sufficiently large $d$. For example, the FAC here seems to suggest it's true for all $d$.
So is this only true for sufficiently large $d$? If so, how can we conclude that the Hilbert polynomial exists, that is, how can we conclude that there is a polynomial $P$ so that $P(d) = \chi(\mathcal{O}(d))$ for all $d$?
If you write out the binomial coefficient you have $$\binom{d+r}{d} = \frac{(d+r)(d+r-1)\ldots (d+1)}{r!}$$
So if you define $$P(d) = \frac{(d+r)(d+r-1)\ldots (d+1)}{r!}$$
It is easy to see that $P(d) = \binom{n+d}{d}$ for $d\ge 0$, in the usual sense. For $d<0$, $P(d)$ still makes sense, you have $P(-1)=\ldots = P(-r) = 0$ and $P(d) = (-1)^r \binom{-d-1}{-n-d-1}$, as you would expect.