Understanding binomial theorem with negative integer of n

Question

I’m a student studying mathematics and I understood how the binomial theorem with positive integer of n works.

But I couldn’t really understand the progress of how the binomial theorem with negative power works. By searching internet, I found the formula of binomial theorem with negative integer of $n$ but I can’t understand how it came across. (Also, why does it doesn’t start with $(a+b)$ nor $(x+y)$, but $(1+x)$?)

I’m not sure how it can be progressed from the binomial theorem with positive integer. I’ve seen some proves using combinatorial methods, but I also couldn’t understand.

Help me to understand how binomial theorem with negative power can be drawn.

Binomial requires 2 ("bi") values. To iterate a positive integer you must include an integer which simplifies to 1. Thus with $1+n$ the constant $1$ is bound to $n$. So we can conclude that for any set of $n$, $1+n$ exists as a subset. We can then say for all $x$ in the Universe $N$, $x|x +1$ — Barb, Jun 07 '22 at 08:19
Does this answer your question? How to use the binomial theorem to calculate binomials with a negative exponent - for negative $m$ we have $$ (a+b)^m = a^m \left( 1 + \frac b a \right)^m = a^m \sum_{k=0}^\infty \binom m k \left(\frac b a \right)^k = \sum_{k=0}^\infty \binom m k a^{m-k} b^k. $$ — Dietrich Burde, Jun 07 '22 at 08:20

Understanding binomial theorem with negative integer of n

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