TITLE: Inductive proof for the Binomial Theorem for rising factorials
QUESTION [8 upvotes]: I want to proove the following equality containing rising factorials
$$(x+y)^\overline{n}\overset{(*)}{=}\sum_{k=0}^n\binom{n}{k}x^\overline{k}y^\overline{n-k}.$$
For $n=1$ this equality is obviously correct. Now I tried to create the transition $n-1\rightarrow n$ this way
$$(x+y)^\overline{n}=(x+y)^\overline{n-1}+\underbrace{\binom{n}{n}x^\overline{n}y^\overline{n-n}}_{x^\overline{n}}$$
$$(x+y)^\overline{n}=\sum_{k=0}^{n-1}\binom{n-1}{k}x^\overline{k}y^\overline{n-k-1}+x^\overline{n}.$$
As I am not that familiar with binomials and such sums, can anyone explain me how transform this equality to achieve a representation like in $(*)$?

REPLY [5 votes]: For anyone who's interested, here's a way to prove the identity that doesn't use induction.
$$\sum_{k=0}^n\binom{n}{k}x^\overline{k}y^\overline{n-k} = n! \sum_{k=0}^n \frac{(-1)^k (-x)^{\underline{k}}}{k!} \frac{(-1)^{n-k}(-y)^{\underline{n-k}}}{(n-k)!}$$
$$= n! (-1)^n\sum_{k=0}^n \binom{-x}{k} \binom{-y}{n-k}$$
$$= n! (-1)^n\binom{-x-y}{n}$$
$$= n! (-1)^n\frac{(-x-y)^{\underline{n}}}{n!}$$
$$= (x+y)^{\overline{n}}$$
The first step converts rising powers to falling powers, the second uses the definition of the binomial coefficients for real-valued upper indexes, the third is the Chu-Vandermonde identity, and the rest is converting back to rising powers.