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Since there are two formulas for combination $$\binom{n}{k}={{n(n-1)...(n-k+1) }\over k!}$$ $$\binom{n}{k}= {n!\over k!(n-k)!}\ $$ Then my question is:

How do I expand when $n=-1$ using second formula?

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    Just to expand on the answers below... Intuitively, the combination function you defined above has the intuition of: given a set of $n$ elements, how many unique ways are there to write $k \leq n$ distinct elements together. Unfortunately, our intuition breaks down when $k$ or $n <0$. However, in Complex Analysis, we have a property that if a function is sufficiently "nice" at even a small regions of points, we can "continue" this function over the entire complex plane. This is where the Gamma function referenced below comes into play... – 2017-02-21
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    As such, the only "correct" way to define your combination formula for $n\in \mathbb{Z}^{-}$ (the negative integers), is through the Gamma function. So, if you reference a derivation for the Gamma function (e.g., http://mathworld.wolfram.com/GammaFunction.html) you will see that $\Gamma(n)=(n-1)!$, and as such, this is why Simply Beautiful Art's follows logically. – 2017-02-21

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When $n<0$ (or for any arbitrary $\alpha$, real or even complex), we instead have ${n\choose k}=\frac{\alpha(\alpha-1)(\alpha-2)...(\alpha-k+1)}{k!}$. (Note that $k$ must still be a nonnegative integer.) Hence, the second formula really only works when $n$ is also a nonnegative integer, and the first formula works in general.

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One defines it via limits and the Gamma function:

$$\binom{-1}k=\lim_{n\to-1}\frac{\Gamma(1+n)}{k!\Gamma(1+n-k)}=\frac{(-1)^{k+1}}{(k!)^2}$$