Intuitive explanation for negative binomial expansion I understand the relationship between Pascal's triangle and binomial expansion for nonnegative integer powers by using combination e.g the binomial coefficient for $x^3$ in $(a + x)^8$ is equivalent to how many ways there are to choose 3 $x$'s from 8 possibilities.
I thought this idea would fail with regards to negative binomial expansion but I find out we can have cases where $r<0$ in $nCr$.
Can someone intuitively explain this? I kind of understand binomial expansion using  Maclaurin series but I can't wrap my head around what negative combination is intuitively and why it works for negative binomial expansion.
Basically I don't understand the reason behind the combination formula still working for negative numbers.
 A: The Question
In this answer, it is shown that
$$
\binom{-n}{j}=(-1)^j\binom{n+j-1}{j}\tag1
$$
If I understand correctly, the question is asking why, when $(1)$ is applied to the Binomial Theorem, we get $(1+x)^{-n}$.
Prior Results
In this answer are three proofs of this cancellation formula
$$
\sum_{j=k}^n(-1)^{j-k}\binom{n}{j}\binom{j}{k}
=[n=k]\tag2
$$
where $[\cdots]$ are Iverson Brackets.

Verification of $\bf{(1)}$ Using the Cauchy Product Formula
For $k\ge1$, we have
$$
\begin{align}
\sum_{j=0}^k\binom{-n}{j}\binom{n}{k-j}
&=\sum_{j=0}^k(-1)^j\binom{n+j-1}{n-1}\binom{n}{n+j-k}\tag{3a}\\
&=\sum_{j=0}^k(-1)^j\sum_{i=0}^{k-1}\binom{n+j-k}{i}\binom{k-1}{n-1-i}\binom{n}{n+j-k}\tag{3b}\\
&=\sum_{i=0}^{k-1}(-1)^{n-k-i}\binom{k-1}{n-1-i}[n=i]\tag{3c}\\[9pt]
&=0\tag{3d}
\end{align}
$$
Explanation:
$\text{(3a)}$: apply $(1)$ and the symmetry of Pascal's Triangle
$\text{(3b)}$: Vandermonde's Identity
$\text{(3c)}$: apply $(2)$
$\text{(3d)}$: $\binom{k-1}{-1}=0$
If $k=0$, the sum is $\binom{-n}{0}\binom{n}{0}=1$. Therefore, using $(1)$, $(2)$, and Vandermonde, we have shown
$$
\sum_{j=0}^k\binom{-n}{j}\binom{n}{k-j}=[k=0]\tag4
$$
which by the Cauchy Product Formula verifies that
$$
\underbrace{\sum_{j=0}^\infty\binom{-n}{j}x^j}_{(1+x)^{-n}}\ \underbrace{\sum_{j=0}^n\binom{n}{j}x^j}_{(1+x)^n}=1\tag5
$$

Verification of $\bf{(1)}$ Using Induction
Using the formula for the sum of a geometric series, we have
$$
(1+x)^{-1}=\sum_{k=0}^\infty(-1)^kx^k\tag6
$$
We can verify $(1)$ inductively using $(6)$ and the Cauchy Product Formula. Assume that $(1)$ is true for a given $n$, then
$$
\begin{align}
(1+x)^{-n-1}
&=\color{#C00}{(1+x)^{-1}}\color{#090}{(1+x)^{-n}}\tag{7a}\\[9pt]
&=\sum_{k=0}^\infty\sum_{j=0}^k\color{#C00}{(-1)^{k-j}x^{k-j}}\color{#090}{(-1)^j\binom{n+j-1}{j}x^j}\tag{7b}\\
&=\sum_{k=0}^\infty\sum_{j=0}^k(-1)^k\binom{n+j-1}{j}x^k\tag{7c}\\
&=\sum_{k=0}^\infty(-1)^k\binom{n+k}{k}x^k\tag{7d}\\
\end{align}
$$
Explanation:
$\text{(7b)}$: $(6)$ and the inductive hypothesis
$\text{(7c)}$: collect terms
$\text{(7d)}$: Hockey-Stick Identity
which verifies $(1)$ for $n+1$.
A: $(1+x)^{-n}$ can also be written as $\frac{1}{(1+x)^n}$.
The denominator can be expanded and then viewed as the usual identity $$\frac1{1-x}=1+x+x^2+\dots$$
For example with $n=2$:
$$1+(-2x-x^2)+(2x+x^2)^2+(-2x-x^2)^3+\dots$$
This can have the like terms collected, which gives:
$$\binom{-2}{1}=-2$$
$$\binom{-2}{2}=3$$
etc...
