As Akiva Weinberger states, the factorial of a negative integer is undefined, however, there are some nifty ways we can get around to negative non-integers... though they don't represent much to my knowledge.
First notice that for positive whole numbers, we have
$$\frac{n!(n+1)^m}{(n+m)!}=\frac{(n+1)^m}{(n+1)\dots(n+m)}=\frac{\left(1+\frac1n\right)^m}{\left(1+\frac1n\right)\dots\left(1+\frac mn\right)}$$
The first step follows by expanding the factorials and cancelling like factors. Then divide the entire fraction by $n^m/n^m$ to get the second step.
As $n$ tends to become infinitely large, then for any $m$, we get
$$\lim_{n\to\infty}\frac{n!(n+1)^m}{(n+m)!}=\lim_{n\to\infty}\frac{\left(1+\frac1n\right)^m}{\left(1+\frac1n\right)\dots\left(1+\frac mn\right)}=\frac{(1+0)^m}{(1+0)\dots(1+0)}=1$$
Where we make use of the fact that $1/\infty=0$ (to be quick and hand-wavy)
Assuming this property holds for non-whole numbers $m$, we now state that the limit
$$\lim_{n\to\infty}\frac{n!(n+1)^m}{(n+m)!}=1$$
We now let non-whole number $x$ take the place of $m$,
$$\lim_{n\to\infty}\frac{n!(n+1)^x}{(n+x)!}=1$$
Multiply both sides by $x!$ to get
$$\lim_{n\to\infty}\frac{n!x!(n+1)^x}{(n+x)!}=x!$$
Notice that in the same fashion of expanding the factorials (assuming non-integer factorials behave like they normally would), we may find that
$$\frac{x!}{(n+x)!}=\frac1{(x+1)\dots(x+n)}$$
And thus,
$$\lim_{n\to\infty}\frac{n!(n+1)^x}{(x+1)\dots(x+n)}=x!$$
WHOA WHOA WHOA! HOLD UP! Do you see what I see?
$$\begin{array}{c|c|c}&\text{makes sense}&\text{makes no sense}\\\hline n!&\color{green}\checkmark\\(n+1)^x&\color{green}\checkmark\\(x+1)\dots(x+n)&\color{green}\checkmark\\x!&&\color{green}\checkmark\end{array}$$
*ahem* so the only thing we don't yet understand is a non-integer factorial, that is, $x!$, but everything else makes sense... so why not use this for non-integer factorials?
$$x!=\lim_{n\to\infty}\frac{n!(n+1)^x}{(x+1)\dots(x+n)}$$
Hm... there was an oddity here though... look what happens at a negative integer:
$$(-1)!=\lim_{n\to\infty}\frac{n!(n+1)^{-1}}{\color{red}{(-1+1)}\dots(-1+n)}$$
Ah... so negative integers result in division by zero. Go figure if you read Akiva Weinberger's answer. But we can do some fun stuff with this, like...
$$(1/2)!=\lim_{n\to\infty}\frac{n!\sqrt{n+1}}{\left(\frac12+1\right)\dots\left(\frac12+n\right)}$$
And if you take a look at this graph, you will see that
$$(1/2)!=\frac{\sqrt\pi}2$$
This extends beyond negative numbers as well. Indeed, you could even take complex numbers into the scheme:
$$i!=\lim_{n\to\infty}\frac{n!(n+1)^i}{(i+1)\dots(i+n)}$$
Other forms of the extended factorial (Gamma function) may be found on Wikipedia:
Gamma function
From Wikipedia, the free encyclopedia
