Expectation for a product of Geometric random variables 
Problem: Let $X$ be Geometric$(p)$. Show for $r = 2,3,4\dots$, that 
  $$E[X(X-1)\dots(X-r+1)]=\frac{r!p^r}{(1-p)^r}$$

Attempt: $P(X=k)=p^k(1-p)$
\begin{align}E[X(X-1)\dots(X-r+1)]&=\sum_{k=0}^\infty k(k-1)\dots(k-r+1)p^k(1-p)\\[0.3cm]^{\text{As the terms when $k < r$ turn out to be $0$.}}&=\sum_{k=r}^\infty k(k-1)\dots(k-r+1)p^k(1-p)
\\[0.3cm]&=r!\sum_{k=r}^\infty \binom{k}{r} p^k(1-p)
\\[0.3cm]&=r!\frac{p^r}{(1-p)^{r-1}}\sum_{k=r}^\infty \binom{k}{r} p^{k-r}(1-p)^{r}\end{align}
I got stuck then, could you please give me some hint on this?
 A: Hint: Change of bound variable: $$\sum_{k=r}^\infty \binom k{k-r}p^{k-r}(1-p)^{r} ~=~ \sum_{n=0}^\infty \binom{n+r}{n}p^n (1-p)^r$$
Which you should (but may not) recognise as the negative binomial series.

 $$\sum_{n=0}^\infty \binom{n+r}n p^n = \frac{1}{(1-p)^{r+1}}$$

A: It is perhaps simpler to observe that the probability generating function defined by $$P_X(t) = \operatorname{E}[t^X]$$ obeys the relationship $$\left[\frac{d^r P_X(t)}{dt^r}\right]_{t=1} = P_X^{(r)}(1) = \operatorname{E}[X(X-1)\ldots (X-r+1)], \quad r = 1, 2, \ldots,$$ whenever such moments are defined.  Then for a geometric random variable with $$\Pr[X = x] = p^x (1-p), \quad x = 0, 1, 2, \ldots,$$ we immediately have $$P_X(t) = \sum_{x=0}^\infty t^x p^x (1-p) = \frac{1-p}{1-pt}, \quad |t| < 1/p.$$  Computing successive derivatives with respect to $t$ leads to the pattern
$$P_X'(t) = p(1-p)(1-pt)^{-2}, \\
P_X''(t) = 2p^2(1-p)(1-pt)^{-3}, \\
P_X'''(t) = 6p^3(1-p)(1-pt)^{-4}, \\
\ldots,$$
hence $$P_X^{(r)}(t) = \frac{r! p^r (1-p)}{(1-pt)^{r+1}},$$ where upon evaluating at $t = 1$ gives the desired result.

Alternatively, we can use the (generalized) binomial theorem and compute the expectation directly, first observing that $$X(X-1)\ldots (X-r+1) = \frac{X!}{(X-r)!} = r! \binom{X}{r}, \quad X \ge r,$$ and $0$ otherwise.  Hence $$\operatorname{E}[X(X-1)\ldots (X-r+1)] = r! (1-p) \sum_{x=r}^\infty \binom{x}{r} p^x,$$ and the result immediately follows.
