I read that the $(r+1)^{\text{th}}$ of the binomial theorem is$$\frac {n(n-1)(n-2)(n-3)\ldots (n-r+1)}{r!}a^{n-r}b^r=\frac {n!}{(n-r)!\cdot r!}a^{n-r}b^r\tag{1}$$Where the binomial theorem is$$(a+b)^n=a^n+na^{n-1}b+\frac {n(n-1)}{2!}a^{n-2}b^2+\frac {n(n-1)(n-2)}{3!}a^{n-3}b^3+\&c\tag{2}$$ Only if $n$ is a positive integer (i.e $n\in\mathbb{Z}^+$). If $n$ is a negative integer, then we have the expansion of $(2)$ as$$(a+b)^{-n}=a^{-n}-na^{-n-1}b+\frac {n(n+1)}{2!}a^{-n-2}b^2+\frac {n(n+1)(n+2)}{3!}a^{-n-r}b^r+\&c\tag{3}$$ With its $(r+1)^{\text{th}}$ term as$$(-1)^r\cdot\frac {n(n+1)(n+2)(n+3)\ldots (n+r-1)}{r!}a^{-n-r}b^r\tag4$$
So my question:
Question: How would we prove $(1)$ and $(4)$ with basic algebra?
I'm not too sure where to even begin; that's how lost I am! I'm thinking my best bet is to use Induction, but I don't know what variable to replace with $x+1$ to make the proof sufficient.
Anything would be helpful at this point! Even hints!