Gamma function $(-1)!$ How would you show that $(-1)!$ is infinite? I think you need to use the gamma function but I'm not entirely sure because I don't think the gamma function works for negative reals
 A: Hint. One may recall that
$$
\int_0^\infty u^se^{-u}\:du=\Gamma(s+1),\quad s>-1. \tag1
$$ Then, integrating by parts, one gets
$$
\Gamma(s+1)=s\:\Gamma(s),\qquad s>0, \tag2
$$ giving
$$
\Gamma(s)=\frac{\Gamma(s+1)}s,\qquad s>0, 
$$ and, as $s \to 0^+$,
$$
\Gamma(s)\sim\frac1s 
$$ yielding
$$
\lim_{s \to 0^+}\Gamma(s)=+\infty
$$ Then, if one defines $'(-1)!'$ as $'\Gamma(0)'=\lim_{s \to 0^+}\Gamma(s)$ , one obtains the announced result.
A: One needs two conditions for the factorial:


*

*$n!=n(n-1)!$

*$1!=1$
So it becomes clear that
$$1!=1\times0!\implies0!=1$$
$$0!=0\times(-1)!\implies(-1)!=1/0\to\pm\infty$$
Anytime a division by $0$ occurs, the factorial (and the gamma function) diverge to $\pm\infty$.  This occurs at the negative integers.
The gamma function, on the other hand, does not diverge over the negative reals (except at the negative integers).  Examine that $\Gamma(-1/2)=-2\sqrt\pi$, for example.
A: Define the following 
$$\Gamma(x+1) = x\Gamma(x)$$
Then we clearly see that 
$$\Gamma(x+1) = x(x-1)\cdots(x-n)\Gamma(x-n)$$
That implies the gamma function diverges for non-positive integers. 
