I'm trying to prove $$\sum_{k=r}^\infty \binom{k-1}{r-1}p^rq^{k-r}= 1$$
I'm given a hint to use $(a+b)^m = \sum_{k=0}^\infty \binom{m}{k} a^kb^{m-k}$.
I choose to let $a = -q$, $b = 1$, $m = -r$ and I get
$$(1-q)^{-r} = \sum_{k=0}^\infty \dfrac{(-r)(-r-1)...(-r-k +1)}{k!}(-q)^k$$
However I am not sure how to transform $$\sum_{k=r}^\infty \binom{k-1}{r-1}p^rq^{k-r} = p^r \sum_{k=r}^\infty \binom{k-1}{r-1}q^{k-r}$$ to something which can use that identity
edit: $q = 1-p$
I don't know how to prove the identity using the binomial theorem and I provide a proof using induction (thus the proof may be longer).
Let $S_r = \sum_{k=r}^{\infty}{k-1 \choose r-1}p^rq^{k-r}$. We only need to prove $\forall r \geq 1, S_r = 1$ since by convention, the binomial coefficient is defined as 0 when the lower index is negative.
basis: $S_1 = \sum_{k=1}^{\infty}{k-1 \choose 0}pq^{k-1} = p\sum_{k=0}^{\infty}q^k=p\cdot\frac{1}{1-q}=1$
induction: Assume $S_r = 1, r \geq 1$.
$ \quad S_{r+1}\\ =\sum_{k=r+1}^{\infty}{k-1 \choose r}p^{r+1}q^{k-r-1}\\ =\sum_{k=r+1}^{\infty}{k-2 \choose r-1}p^{r+1}q^{k-r-1} + \sum_{k=r+1}^{\infty}{k-2 \choose r}p^{r+1}q^{k-r-1}\\ =\sum_{k=r}^{\infty}{k-1 \choose r-1}p^{r+1}q^{k-r} + \sum_{k=r+1}^{\infty}{k-2 \choose r}p^{r+1}q^{k-r-1}\\ =pS_r+ \sum_{k=r+1}^{\infty}{k-2 \choose r}p^{r+1}q^{k-r-1}\\ =pS_r+\sum_{k=r}^{\infty}{k-1 \choose r}p^{r+1}q^{k-r}\\ =pS_r+q(S_{r+1}+{r-1\choose r}p^{r+1}q^{-1})\\ =pS_r+qS_{r+1}=p+qS_{r+1}$
thus $S_{r+1}=1$. QED.
The second equality comes from the addition identity of binomial coefficients, say, ${r \choose k}={r-1 \choose k}+{r-1\choose k-1}$.
First, let's figure out what happens when we negate the upper index on a binomial coefficient: $$\begin{align*} \binom{-a}{b} &= \frac{(-a)!}{b! \, (-a-b)!} \\ &= \frac{(-a)(-a-1)\cdots(-a-b+1)(-a-b)!}{b!(-a-b)!} \\ &= \frac{(-a)(-a-1)\cdots(-a-b+1)}{b!} \\ &= (-1)^b \frac{a(a+1)\cdots(a+b-1)}{b!} \\ &= (-1)^b \frac{(a+b-1)!}{(a-1)! \, b!} \\ &= (-1)^b \binom{a+b-1}{a-1}.\end{align*}$$ Note that the sum we want to evaluate is a summation over the upper index of the binomial coefficient, whereas the binomial theorem we are supposed to use is a summation over the lower index. But the identity we proved above does precisely this conversion: if we sum over $b$, the first expression has $b$ on the bottom, but the last expression has $b$ on the top only. So all we need is to choose the variables appropriately. But first, we shift the index of summation to zero by letting $j = k - r$, hence $k = r+j$:
$$\begin{align*} \sum_{k=r}^\infty \binom{k-1}{r-1} p^r (1-p)^{k-r} &= \sum_{j=0}^\infty \binom{r+j-1}{r-1} p^r (1-p)^j \\ &= \sum_{j=0}^\infty (-1)^{-j} \binom{-r}{j} p^r (1-p)^j \\ &= p^r \sum_{j=0}^\infty \binom{-r}{j} (p-1)^j 1^{r-j} \\ &= p^r ((p-1)+1)^{-r} \\ &= p^r p^{-r} \\ &= 1. \end{align*}$$
\begin{eqnarray*} \sum_{k=r}^{\infty}{\binom{k-1}{r-1}p^rq^{k-r}} &=& \sum_{k=0}^{\infty}{\binom{k+r-1}{k}p^rq^k} \qquad\qquad\qquad\text{replacing $k$ with $k+r$} \\ &=& \sum_{k=0}^{\infty}{\binom{-r}{k}p^r(-q)^k} \qquad\text{using identity $\binom{k+r-1}{k}=\binom{-r}{k}(-1)^k$} \\ && \qquad\qquad\qquad\text{(easily verified from definition of binomial coefficient)} \\ &=& p^r \sum_{k=0}^{\infty}{\binom{-r}{k}(-q)^k} \\ &=& p^r (1-q)^{-r} \qquad\qquad\text{using the hint with $a=1,b=-q,m=-r$} \\ &=& 1. \end{eqnarray*}
