The binomial identity: $(x + y)^n = \sum_{k = 0}^{n}{n \choose k}x^{n - k}y^{k}$
seems closely related to: $$(x + y)_n = \sum_{k = 0}^{n}{n \choose k}(x_{n - k})(y_{k})$$ For $$(x)_{n} = (x) * (x- 1) * \dots * (x- n + 1), (x)_0 = 1$$ I'm trying to prove the second identity, for all $x \in \mathbb{C}$, so the standard induction proof does not apply. For positive $x, y$, a simple combinatorial argument should work, but I can't generalize to all $x \in \mathbb{C}$.
As in this answer, in general we have $$ \frac{d^n}{dt^n} f(t)g(t) = \sum_{k=0}^n \binom{n}{k} \left( \frac{d^k}{dt^k} f(t) \right) \left( \frac{d^{n-k}}{dt^{n-k}} g(t) \right). $$ (The proof is by induction on $n$, using the product rule and Pascal's identity.)
Now put $f(t) = t^y$ and $g(t) = t^x$. Then the two sides of the above equation are $$ \frac{d^n}{dt^n} t^y t^x = \frac{d^n}{dt^n} t^{x+y} = (x+y)_n t^{x+y-n},$$ and \begin{align} \sum_{k=0}^n \binom{n}{k} \left( \frac{d^k}{dt^k} t^y \right) \left( \frac{d^{n-k}}{dt^{n-k}} t^x \right) &= \sum_{k=0}^n \binom{n}{k} (y)_k t^{y-n+k} \cdot (x)_{n-k} t^{x-k} \\ &= \sum_{k=0}^n \binom{n}{k} (x)_k (y)_{n-k} t^{x+y-n}. \end{align} Putting $t = 1$ gives the desired result.
Note that if we divide both sides of your equation by $n!$ then (recalling that $\binom{m}{n} = (m)_n/n!$) we end up with Vandermonde's identity $$ \binom{x+y}{n} = \sum_{k=0}^n \binom{x}{n-k} \binom{y}{k}. $$
Here's another proof that is more general. Let $f(t)$ be any power series with $f(0) = 0$. Define a polynomial sequence $p_n(x)$ by $$\sum_{n=0}^\infty \frac{p_n(x)}{n!} t^n = e^{tf(x)}.$$
Polynomials $p_n(t)$ of this form are said to be of binomial type because they obey a binomial theorem:
Claim: We have $p_n(x+y) = \sum_{k=0}^\infty {n \choose k} p_k(x) p_{n-k}(x).$
Proof: Consider $e^{(x+y)f(x)}$. We have $$e^{(x+y)f(t)} = \sum_{n=0}^\infty \frac{p_n(x+y)}{n!} t^n.$$ On the other hand, \begin{align*} e^{xf(t)} e^{yf(t)} &= \left(\sum_{i=0}^\infty \frac{p_i(x)}{i!} t^i\right) \left(\sum_{j=0}^\infty \frac{p_j(x)}{j!} t^j\right)\\ &= \sum_{n=0}^\infty \sum_{i+j=n}\frac{p_i(x)}{i!} \frac{p_j(x)}{j!} t^n\\ &= \sum_{n=0}^\infty \sum_{k=0}^n {n \choose k} p_k(x)p_{n-k}(y) \frac{t^n}{n!}. \end{align*}
Now comparing the coefficient of $t^n/n!$ in the two equations proves the claim.
To get the falling factorials, take $f(t) = \log(1+t)$. Then $$e^{xf(t)} = e^{x \log(1+t)} = (1+t)^x = \sum_{n=0}^\infty {x \choose n} t^n = \sum_{n=0}^\infty \frac{(x)_n}{n!} t^n$$ so $p_n(x) = (x)_n$. To get the usual binomial theorem, take $f(t) = t$.
