Why does $\sum_{k\geq0}\binom{-r}{k}p^r(p-1)^k=p^r(1+p-1)^{-r}$?

For any positive real number $r$, it is clear that $\binom{-r}{k}(-1)^k\geq0$ for all positive integer $k$. The general binomial theorem then implies

$$\sum_{k\geq0}\binom{-r}{k}p^r(p-1)^k=p^r(1+p-1)^{-r}=1$$

I get the generalisation of $\binom{-r}{k}$, and I have searched through Wikipedia about the Binomial Theorem, it only mentions the generalisation to multinomial theorem and complex numbers.

I don't quite get how do they get $p^r(1+p-1)^{-r}$. It is very appreciated if anyone could explain a bit on it. Thanks!

• – lab bhattacharjee Mar 25 '16 at 11:39
• @labbhattacharjee What part of the link you gave can aid the OP in her/his quest? – complexmanifold Mar 25 '16 at 11:41

I imagine this will give you the answer you need. In particular, observe that $p^r$ can be factored out, so you end up with $$p^r\sum_{k \geq 0}\binom{-r}{k}(p-1)^k = p^r\sum_{k \geq 0}\binom{-r}{k}(p-1)^k(1)^{-r-k} = p^r(1+p-1)^{-r} = 1\text{.}$$