Let X be a geometric random variable, show that $E[X(X-1)...(X-r+1)] = \frac{r!(1-p)^r}{p^r}$ 
Let $X$ be a geometric random variable, i.e., $P(X=i)=(1-p)^ip$ for $i=0,1,2,...$ where $0<p<1$. Show that $$E[X(X-1)...(X-r+1)] = \frac{r!(1-p)^r}{p^r}$$ 

I try to solve it like this:
\begin{align}
&E[X(X-1)...(X-r+1)]\\
=&\sum_{i=0}^{\infty}i(i-1)...(i-r+1)\cdot(1-p)^ip\\
=&\sum_{i=0}^{\infty}\frac{i!}{(i-r)!}\cdot(1-p)^ip\\
=&r!p\sum_{i=0}^{\infty}\frac{i!}{(i-r)!r!}(1-p)^i
\end{align}
and then I don't know how to continue.
 A: I think it is hard to complete your argument. Use of calculus will yield the result fairly easily: for $0\leq t \leq 1$ let $f(t)=Et^{X}=\sum_0 ^{\infty} t^{i} (1-p)^{i} p$. Clearly, the sum here is a geometric sum so we can compute it explicitly to get $f(t)=\frac p {1-t(1-p)}$. Now differentiate this $r$ times to get $\sum X(X-1)...(X-r+1)t^{X-r}$. If you put $t=1$ in this you get $EX(X-1)...(X-r+1)$. So the value of this expectation is $\frac {d^{r}} {dt^{r}} f(t)$ evaluated at $t=1$.  It is fairly easy to calculate this derivative.
A: If you want to continue it that way, you need a version of the binomial theorem telling you that $\sum_{i=0}^\infty \binom{i+r}{i}z^i = (1-z)^{-r-1}$ --- see here for example.
Your sum is $r! p \sum_{i\geq 0} \binom{i}{i-r} (1-p)^i$ which you can rewrite as $(1-p)^r r! p \sum_{i\geq 0} \binom{i+r}{i}(1-p)^i$ by noting that the terms with $i<r$ in your sum are zero. The binomial theorem tells you this is $(1-p)^r r! p (1-(1-p))^{-r-1}$ which is what you wanted.
