Quotient of Gamma functions I am trying to find a clever way to compute the quotient of two gamma functions whose inputs differ by some integer. In other words, for some real value $x$ and an integer $n < x$, I want to find a way to compute
$$ \frac{\Gamma(x)}{\Gamma(x-n)} $$
For $n=1$, the quotient it is simply $(x-1)$ since by definition
$$ \Gamma(x) = (x - 1)\Gamma(x-1) $$
For $n=2$, it is also simple:
$$ \frac{\Gamma(x)}{\Gamma(x-2)} = 
\frac{(x-1)\Gamma(x-1)}{\Gamma(x-2)} =
(x-1)(x-2)$$
If we continue this pattern out to some arbitrary $n$, we get
$$ \frac{\Gamma(x)}{\Gamma(x-n)} = \prod_i^n (x-i)$$
Obviously I am floundering a bit here. Can anyone help me find an efficient way to compute this quotient besides directly computing the two gamma functions and dividing?
I am also okay if an efficient computation can be found in log space. Currently I am using a simple approximation of the log gamma function and taking the difference. This was necessary because the gamma function gets too big to store in any primitive data type for even smallish values of $x$.

I think you mean
$$ \frac{\Gamma(x)}{\Gamma(x-n)} = \prod_{i=1}^{n} (x - i) $$
Of course this might not be very nice if $n$ is very large, in which case you might want to first compute the $\Gamma$ values and divide; but then (unless $x$ is very close to one of the integers $i$) the result will also be enormous.  If you're looking for a numerical approximation rather than an exact value, you can use Stirling's approximation
or its variants.
EDIT: Note also that if your $x$ or $x-n$ might be negative, you may find the reflection formula useful:
$$ \Gamma(z) \Gamma(1-z) = \frac{\pi}{\sin(\pi z)} $$ 
For $x > n$, 
$$ \eqalign{\ln\left(\frac{\Gamma(x)}{\Gamma(x-n)}\right) &= n \ln x + \sum_{i=1}^n \ln(x-i)\cr
&= n \ln x + \sum_{i=1}^n \left( - \frac{i}{x} - \frac{i^2}{2x^2} - \frac{i^3}{3x^3} - \frac{i^4}{4x^4} - \frac{i^5}{5x^5} - \frac{i^6}{6 x^6}\ldots \right)\cr
&= n \ln x - \frac{n(n+1)}{2x} - \frac{n(n+1)(2n+1)}{12 x^2} - \frac{n^2 (n+1)^2}{12 x^3} \cr
& -{\frac {n \left( n+
1 \right)  \left( 2\,n+1 \right)  \left( 3\,{n}^{2}+3\,n-1 \right) }{120 {
x}^{4}}}-\frac { \left( n+1 \right) ^{2}{n}^{2}
 \left( 2\,{n}^{2}+2\,n-1 \right) }{60 {x}^{5}}
\ldots} $$
This provides excellent approximations as $x \to \infty$ for fixed $n$ (or $n$ growing much more slowly than $x$).
On the other hand, for $n = tx$ with $0 < t < 1$ fixed, as $x \to \infty$ we have
$$ \eqalign{\ln\left(\frac{\Gamma(x)}{\Gamma(x-n)}\right) &=(1-t) x \ln(x) - (t \ln(t) - t + 1) x +\frac{\ln(t)}{2}\cr &+{
\frac {t-1}{12tx}}+{\frac {1-{t}^{3}}{360{t}^{3}{x}^{3}}
}+{\frac {{t}^{5}-1}{1260{t}^{5}{x}^{5}}}+\frac{1-t^7}{1680 t^7 x^7} \ldots 
}$$
