How $\alpha(\alpha+1)\ldots(\alpha+k-1)=\frac{\Gamma(\alpha+k)}{\Gamma(\alpha)}$?

Question

Probability function of Negative Binomial Distribution, $NB(\alpha,p)$, is

$$P(X=k)=\binom{\alpha+k-1}{k}(1-p)^{\alpha}p^k,\quad \alpha>0$$

Probability generating function of Negative Binomial Distribution is

$$P_X(s)=(\frac{1-p}{1-ps})^{\alpha}$$

Now i have to compute the $k^{th}$ Factorial Moment,$\mu_k$.

I know that if i differentiate $P_X(s)$ $k$-times with respect to $s$ and put $s=1$ then i will get the $k^{th}$ Factorial Moment,$\mu_k$.

My result is $$\mu_k=P^{(k)}_X(1)=\alpha(\alpha+1)\ldots(\alpha+k-1)\frac{p^k}{(1-p)^{k}}$$

But in the book the result is :

$$\mu_k=\frac{\Gamma(\alpha+k)}{\Gamma(\alpha)}\times\frac{p^k}{(1-p)^{k}}$$

How $$\alpha(\alpha+1)\ldots(\alpha+k-1)=\frac{\Gamma(\alpha+k)}{\Gamma(\alpha)}$$?

Answer 1

We know that $\Gamma(z+1) = z\,\Gamma(z)$ (for $z\in\mathbb{C}\setminus\mathbb{Z}_{\le 0}$). Now looking at $\Gamma(z+2)$ we have $\Gamma(z+2) = (z+1)\Gamma(z+1) = (z+1)z\,\Gamma(z)$. Given that, it is easy to see with an inductive argument that

$$\Gamma(\alpha + k) = (\alpha+k-1)(\alpha+k-2)...\alpha\,\Gamma(\alpha).$$

Dividing both sides by $\Gamma(\alpha)$ gives you your answer. This is often called the Pochhammer symbol.

Answer 2

Use that $\Gamma(1+x)=x\Gamma(x)$, $k$ times on $\Gamma(\alpha+k)$.