Coefficeient of $x^k$ in $(1+x)^n$ when $n<0$

Question

I know this is a very basic question. But I simply cannot derive the final answer. We have the alternate form of binomial theorem if we want to deal with negative exponents. $$(1+x)^n=1+nx+\frac{n(n-1)}{2!}x^2+\cdots $$ If $n<0$ the coefficient of $x^k$ is given by $n+k-1 \choose k$. I am trying to derive it from the above series. The coefficient of $x^k$ is $$\frac{n(n-1)(n-2)\cdots (n-(k-1))}{k!}\tag{1}$$ I have no idea or intuition on how to proceed from (1) to $n+k-1 \choose k$ when $n<0$.

Answer 1

Use the binomial theorem:

$$(1+x)^n=\sum_{i=1}^n\binom nix^i\implies\;\text{the coefficient is}\;\;\binom nk:=\frac{n!}{k!(n-k)!}$$