Expanding $\frac{\Gamma(n)}{\Gamma(n-k)}$ as a polynomial I want to expand $\frac{\Gamma(n)}{\Gamma(n-k)}$ as a polynomial, where $\Gamma$ is the gamma function.
For $k\in\mathbb{N}$, it can be "simplified" as
$$\frac{\Gamma(n)}{\Gamma(n-k)}=(n-1)(n-2)(n-3)\dots(n-k)$$
I was wondering if it were possible to expand that into $\sum_{i=0}^ka_in^{k-i}$ form.
Then, there is the harder case of $k\in\mathbb R$.
I imagined that it would be of the form $\Pi_{i=0}^\infty(n-r_i)$ where $r_n$ is the $n$th root.
If this were the case, we see that
$$\Gamma(n)\ne0\implies\frac1{\Gamma(n-k)}=0$$
This occurs at the poles
$$\implies r_i-k=-i$$
$$r_i=i+k$$
And put into product form
$$\frac{\Gamma(n)}{\Gamma(n-k)}=\Pi_{i=0}^\infty(n-r_i)=\Pi_{i=0}^\infty(n-k-i)$$
Sadly, I don't think it is possible to multiple the product out because it diverges.


*

*Q1 Is there a polynomial/expanded form for $k\in\mathbb N$ using summations if needed?

*Q2 What about $k\in\mathbb R$?
EDIT
From below, you can see I have worked the case for $k\in\mathbb N$, but I still need $k\in\mathbb R$.
$$\frac{\Gamma(u)}{\Gamma(u-n)}=\sum_{k=0}^n(-1)^{n-k}s(n,k)(u-n)^k$$
My assumption for $n\in\mathbb R$ is that the formula becomes
$$\frac{\Gamma(u)}{\Gamma(u-n)}=\sum_{k=0}^\infty(-1)^{n-k}s(n,k)(u-n)^k$$
For similar reasons for why Euler extended binomial expansion the way he did.  I note that my proposed formula holds true if $n\in\mathbb N$, but doesn't make much sense for $n\in\mathbb R$.
 A: As WolframAlpha states:
$$\frac{\Gamma(x+n)}{\Gamma(x)}=x(x+1)(x+2)(x+3)\dots(x+n-1)=\sum_{k=0}^n(-1)^{n-k}s(n,k)x^k$$
where $s(n,k)$ is a Stirling number of the first kind.
Rewritable as
$$x(x+1)(x+2)(x+3)\dots(x+n-1)=(x+n-1)(x+n-2)(x+n-3)\dots x=\frac{\Gamma(x+n)}{\Gamma(x+n-n)}$$
Use substitution $x+n=u$
$$=\frac{\Gamma(u)}{\Gamma(u-n)}$$
A: Since $\frac{\Gamma(x+n)}{\Gamma(x)}=(x)_n$, the answer is simply given by the generating function for Stirling numbers of the first kind.
A: In particular, for $\displaystyle k=\frac{1}{2}$,
\begin{align*}
  \Gamma \left( n-\frac{1}{2} \right) &= \frac{(2n-3)!!}{2^{n-1}} \sqrt{\pi} \\
  \frac{\Gamma \left( n \right)}{\Gamma \left( n-\frac{1}{2} \right)} &=
  \frac{(n-1)!}{(2n-3)!!} \frac{2^{n-1}}{\sqrt{\pi}} \\
  &= \frac{(n-1)!(2n-2)!!}{(2n-2)!!(2n-3)!!} \frac{2^{n-1}}{\sqrt{\pi}} \\
  &= \frac{2^{n-1} [(n-1)!]^{2}}{(2n-2)!} \frac{2^{n-1}}{\sqrt{\pi}} \\
  &= \frac{2^{2(n-1)}}{\binom{2n-2}{n-1}\sqrt{\pi}} \\
  &= \frac{2^{2(n-1)}(n-1)!}{(2n-2)(2n-3) \ldots n \sqrt{\pi}}
\end{align*}
which is not a polynomial in $n$.
By $\displaystyle \Gamma \left( \frac{1}{3} \right)=
2^{7/9}3^{-1/12} \sqrt[3]{\pi K(\sin 15^{\circ})}$, we can do the something similar for $k=\frac{1}{3}$.
