Generalizing binomial coefficient to negative integers

Generatingfunctionology claims that $\binom{n}{k} = 0$  for $k < 0$. I have checked using formula $\binom{n}{k} = {n! \over k! (n-k)!}$ and the fact that $$0! = 1 = (-1)! * 0\ \ \text{ identical to }\ \ (-1)! = {1 \over 0}$$ $$-1! = (-2)! * (-1)\ \ \text{ identical to }\ \ (-2)! = {1 \over 0 (-1)}$$ $$-2! = (-3)! * (-2)\ \ \text{ identical to }\ \ (-3)! = {1 \over 0 (-1)(-2)}$$ and, generally, $$(-n)! = {1 \over 0 (-1)(-2) \cdots (-n+1)}$$

I have got the following Karnaugh map, which shows whether $\binom{n}{k} = 0$ or not,

		n > 0
	=	≠	=	≠
n>k	≠	=	≠	=
			k > 0

You see a checkerboard order of =0 and ≠0: every time we switch one of the conditions, n>k, n>0 or k>0, the output $\binom{n}{k}$ switches between zero to non-zero. The same checkerboard behaviour is also observed in Wikipedia. Now, we see that author's statement that "$\binom{n}{k} = 0$ vanishes unless 0 ≤ k ≤ n" is true only for $n > 0$.