TITLE: Negative Binomial Coefficients
QUESTION [6 upvotes]: Is it true that Pascal's Rule holds for binomial coefficients with a negative upper index?  
With $n = -1$ and $k = 3$, for example, it appears not to hold.

REPLY [4 votes]: The binomial coefficients may actually be viewed as rational-coefficient polynomials:
$${X\choose k}:=\frac{X(X-1)\cdots\big(X-(k-1)\big)}{k\cdot(k-1)\cdots3\cdot2\cdot1}.$$
This is useful/interesting for a number of reasons. These polynomials are a basis for the space of all rational-coefficient polynomials which take integer values at integer arguments. The binomial theorem generalizes to powers that need not be natural numbers:
$$(1+T)^s=\sum_{k\ge0}{s\choose k}T^k.$$
This allows us to extend the notion of taking-powers to complete topological rings of characteristic zero (which is to say, we can do it in abstract algebraic structures equipped with limits where a priori it is not defined), although there are caveats as to regions of convergence. It may also be useful in the contexts of generating functions and special functions.
Anyway, Pascal's rule holds on the more fundamental level of polynomials:
$${X-1\choose k}+{X-1\choose k-1}=\frac{\color{Green}{(X-1)\cdots}\color{Blue}{\big(X-1-(k-1)\big)}}{k!}+\frac{\color{Purple}{k}\color{Green}{(X-1)\cdots\big((X-1)-(k-2)\big)}}{k!}$$
$$=\frac{\color{Green}{(X-1)\cdots (X-(k-1))\,}[\color{Blue}{(X-k)}+\color{Purple}{k}]}{k!} ={X\choose k}.$$
Since the identity holds for polynomials, it holds for the polynomials evaluated at any arguments; they can be negative, fractional, irrational, complex, $p$-adic, whatever floats your boat.