On Hilbert polynomials The sequence of Hilbert polynomials is defined by :
$$H_0=1,\ \\ H_n(X)=\frac{X(X-1)...(X-n+1)}{n!}\ ,n>0$$
How can we prove the following identity :
$$H_n(X+Y)=\sum_{k=0}^nH_k(X)H_{n-k}(Y)$$
 A: A little context: This is Vandermonde's identity. Consider the notation $$H_n(X)=\frac{X^{\underline{n}}}{n!},$$ where $X^{\underline{n}}=X(X-1)\cdots (X-n+1).$ Essentially, $H_n(X)=\binom{X}{n}$ if you see it in that way. So what you are proving is $$\binom{X+Y}{n}=\sum _{k=0}^n\binom{X}{k}\binom{Y}{n-k}.$$
Hint: What if you show it for integers(perhaps using binomial theorem $(1+z)^{x+y}=(1+z)^x(1+z)^y$ and expanding for $z.$). Try to extend it for rationals and use approximation to extend it for reals.
A: $$ \left(\forall k\in\mathbb{N}^{*}\right),\ H_{k}\left(X\right)=\frac{1}{k!}\prod_{j=0}^{k-1}{\left(X-j\right)} $$
Let's prove by induction that : $$ \left(\forall k\in\mathbb{N}\right),\ \prod_{j=0}^{k-1}{\left(X+Y-j\right)}=\sum_{j=0}^{k}{\binom{k}{j}\prod_{i=0}^{j-1}{\left(X-i\right)}\prod_{i=0}^{k-j-1}{\left(Y-i\right)}} $$
The statement holds for $ k=1 \cdot $
Let $ n $ be a positive integer, assuming that the statement holds for $ k=n $, let's prove that the statement holds for $ k=n+1 : $
\begin{aligned} \small\prod_{k=0}^{n}{\left(X+Y-k\right)}&\small=\left(X+Y-n\right)\prod_{k=0}^{n-1}{\left(X+Y-k\right)}\\ &\small=\left(X+Y-n\right)\sum_{k=0}^{n}{\binom{n}{k}\prod_{i=0}^{k-1}{\left(X-i\right)}\prod_{i=0}^{n-k-1}{\left(Y-i\right)}}\\ &\small=\sum_{k=0}^{n}{\binom{n}{k}\left(X-k\right)\prod_{i=0}^{k-1}{\left(X-i\right)}\prod_{i=0}^{n-k-1}{\left(Y-i\right)}}+\sum_{k=0}^{n}{\binom{n}{k}\prod_{i=0}^{k-1}{\left(X-i\right)}\left(Y-n+k\right)\prod_{i=0}^{n-k-1}{\left(Y-i\right)}}\\ &\small=\sum_{k=0}^{n}{\binom{n}{k}\prod_{i=0}^{k}{\left(X-i\right)}\prod_{i=0}^{n-k-1}{\left(Y-i\right)}}+\sum_{k=0}^{n}{\binom{n}{k}\prod_{i=0}^{k-1}{\left(X-i\right)}\prod_{i=0}^{n-k}{\left(Y-i\right)}} \\ &\small=\prod_{i=0}^{n}{\left(X-i\right)}+\sum_{k=0}^{n-1}{\binom{n}{k}\prod_{i=0}^{k}{\left(X-i\right)}\prod_{i=0}^{n-k-1}{\left(Y-i\right)}}+\sum_{k=1}^{n}{\binom{n}{k}\prod_{i=0}^{k-1}{\left(X-i\right)}\prod_{i=0}^{n-k}{\left(Y-i\right)}}+\prod_{i=0}^{n}{\left(Y-i\right)}\\ &\small=\prod_{i=0}^{n}{\left(X-i\right)}+\sum_{k=1}^{n}{\binom{n}{k-1}\prod_{i=0}^{k-1}{\left(X-i\right)}\prod_{i=0}^{n-k}{\left(Y-i\right)}}+\sum_{k=1}^{n}{\binom{n}{k}\prod_{i=0}^{k-1}{\left(X-i\right)}\prod_{i=0}^{n-k}{\left(Y-i\right)}}+\prod_{i=0}^{n}{\left(Y-i\right)}\\ &\small=\prod_{i=0}^{n}{\left(X-i\right)}+\sum_{k=1}^{n}{\left[\binom{n}{k}+\binom{n}{k-1}\right]\prod_{i=0}^{k-1}{\left(X-i\right)}\prod_{i=0}^{n-k}{\left(Y-i\right)}}+\prod_{i=0}^{n}{\left(Y-i\right)}\\ &\small=\prod_{i=0}^{n}{\left(X-i\right)}+\sum_{k=1}^{n}{\binom{n+1}{k}\prod_{i=0}^{k-1}{\left(X-i\right)}\prod_{i=0}^{n-k}{\left(Y-i\right)}}+\prod_{i=0}^{n}{\left(Y-i\right)}\\ \small\prod_{k=0}^{n}{\left(X+Y-k\right)}&\small=\sum_{k=0}^{n+1}{\binom{n+1}{k}\prod_{i=0}^{k-1}{\left(X-i\right)}\prod_{i=0}^{n-k}{\left(Y-i\right)}}\end{aligned}
Hence : $$ \left(\forall n\in\mathbb{N}\right),\ \prod_{k=0}^{n-1}{\left(X+Y-k\right)}=\sum_{k=0}^{n}{\binom{n}{k}\prod_{i=0}^{k-1}{\left(X-i\right)}\prod_{i=0}^{n-k-1}{\left(Y-i\right)}} $$
Dividing both sides by $ n! $, we get that : $$ \left(\forall n\in\mathbb{N}\right),\ H_{n}\left(X+Y\right)=\sum_{k=0}^{n-1}{H_{k}\left(X\right)H_{n-k}\left(Y\right)} $$
