Negative Exponents in Binomial Theorem

The below is too long for a comment so I'm including it here even though I'm not sure it "answers" the question.

If you think about $(1+x)^{-n}$ as living in the ring of formal power series $\mathbb{Z}[[x]]$, then you can show that $$(1+x)^{-n} = \sum_{k=0}^{\infty} (-1)^k \binom{n+k-1}{k} x^k$$ and the identity $\binom{-n}{k} = (-1)^k \binom{n+k-1}{k}$ seems very natural. Here's how...

First expand $(1+x)^{-n} = \bigg(\frac{1}{1-(-x)}\bigg)^n = (1 - x + x^2 - x^3 + \dots)^n$. Now, the coefficient on $x^k$ in that product is simply the number of ways to write $k$ as a sum of $n$ nonnegative numbers. That set of sums is in bijection to the set of diagrams with $k$ stars with $n-1$ bars among them. (For example, suppose $k=9$ and $n=4$. Then, **|*|***|*** corresponds to the sum $9=2+1+3+3$; ****||***|**corresponds to the sum $9 = 4+0+3+2$; ****|***||** corresponds to $9=4+3+0+2$; etc.) In each of these stars-and-bars diagrams we have $n+k-1$ objects, and we choose which ones are the $k$ stars in $\binom{n+k-1}{k}$ many ways. The $(-1)^k$ term comes from the alternating signs, and that proves the sum.