Factorial Moment of the Geometric Distribution I am trying to caclulate the Factorial Moment of the Geometric Distribution #2 with parameter $p$. Therefore I set $\Omega = \mathbb{N}_0$ and have by using the pochhammer symbol and setting $q=1-q$ that
$$E((k)_l)= \sum _{k=0}^{\infty } (k)_l p q^k = p^{-l} q \cdot l! \sum _{k=0}^{\infty } (\frac{(k+l-1)!}{(k-1)! \cdot l!}\cdot p^{l+1} q^{k-1}) $$
Now Mathematica tells me that $\sum _{k=0}^{\infty } (\frac{(k+l-1)!}{(k-1)! \cdot l!}\cdot p^{l+1} q^{k-1})=1$, but I cannot see why this identity is true. Also when using 
FactorialMoment[GeometricDistribution[p], l]

Mathematica suggests that $E((k)_l)=(\frac{q}{p})^l l!$. Thank you in Advance for your help.
 A: Let $X$ have geometric distribution, where $X$ is the number of failures before the first success.
The easiest approach to the factorial moments in this case is to find the factorial moment generating function, which is 
$$E(t^X)$$
Suppose the probability of success is $p$.
We want 
$$\sum_{n=0}^\infty pq^n t^n$$
where as usual $q=1-p$. So we want
$$\sum_{n=0}^\infty p(qt)^n$$
Sum this infinite geometric series.
We get
$$\frac{p}{1-qt}$$
To find the $k$-th factorial moment, find the $k$-th derivative of the factorial moment generating function (with respect to $t$) at $t=1$.
In our particular case, finding the $k$-th derivative is easy.
If by geometric distribution you mean total number of trials until first success (so values are $1$, $2$, and so on) a small modification of the above calculation will give the answer.
Addendum: The easiest way to find the sum 
$$\sum_{k=1}^\infty (k)(k-1)\cdots(n-\ell+1)x^k$$
that was asked about is to express this as
$$x^{\ell}\sum_{k=1}^\infty (k)(k-1)\cdots(n-\ell+1)x^{k-\ell}$$
and observe that 
$$(k)(k-1)\cdots(n-\ell+1)x^{k-\ell}$$
is the $\ell$-th derivative of $x^k$.  So the desired sum is the $\ell$-th derivative of $1+x+x^2+ x^3+\cdots$, that is, of $1/(1-x)$.
A: The factorial moments of an integer valued random variable $X$ are linked to the successive derivatives of the generating function $g_X$ of $X$, defined by
$$
g_X(s)=E(s^X)=\sum_{n=0}^{+\infty}P(X=n)s^n.
$$
For every $k\ge0$, the $k$th derivative is
$$
g_X^{(k)}(s)=E((X)_ks^{X-k}),
$$
hence the value at $s=1$ yields the factorial moment.
Now, what is $g_X$ for $X$ geometric?
