How to evaluate binomial coefficients when $k=0$ and $1\geq|n|\geq0$ So I am doing some taylor series which boil down to the geometric series, for which I then need to evaluate various binomial coefficients. I have always used $n\choose k$ $=\frac{n!}{(n-k)!k!}$ to do these, this however becomes troublesome for non-integer and/or negative values of n. 
Now I just memorized that if $k=0$, $n\choose k$ $= 1$ for integers $ n\geq 0$. But this still leaves me with negative and/or negative values for n. My textbook provides me with $n\choose k$ = $\frac{n(n-1)...(n-k+1)}{k!}$, but I`m still not sure how to evaluate this with $k=0$. Somehow I always get it wrong.
For example i get $-1\choose 0$ $=\frac{-1(-1-0+1)}{0!}=\frac{-1*0}{1}=0$, but it should be $1$. Whats my error? Thanks!
 A: For $\binom{-1}{0}$ take $\binom{\alpha}{0}=\frac{\alpha!}{(\alpha-0)!0!}=\frac{1}{1 \cdot 0!}=1$. Along these lines, e.g. $\binom{-1}{10}=\frac{\alpha}{10}=\frac{1}{10!}=\cdots (-1)^{10}\frac{10!}{10!}=1, \ \binom{-1}{11}=(-1)^{11}=-1$, etc
A: I think you got confused by the notation $n(n-1)\dots(n-k+1)$.  This is intended to mean the product of all the numbers you get when you start at $n$ and decrease, in steps of $1$ (to $n-1$ and so forth) until you reach $n-k+1$; so you don't include the factor that would come next, $n-k$.  So there are $k$ factors here, the largest of which is $n$.  Now when $k=0$, you're supposed to still use the same interpretation, not (as the notation made you reasonably think) start counting upward from $n$ instead of downward.  So you should form a product of a sequence of numbers, starting at $n$ and decreasing in steps of $1$ down to and including $n+1$ (that's $n-k+1$) but not including $n$ (that's $n-k$).  So you're supposed to stop before you start!  In other words, when $k=0$, the expression $n(n-1)\dots(n-k+1)$ is intended to mean the product of no factors, which is $1$, not the product of two factors as you supposed.
