The factorial function is defined as
$$n!=\prod_{k=1}^{n}k$$
For $n>0$, $n$ being an integer. This definition can be extended to the complex numbers with positive real part by the Gamma function:
$$\Gamma(n)=(n-1)!$$
This is the plot of the factorial function, in which the values of the factorial of non-integral numbers are evaluated by the gamma function:
It can be seen that for the positive numbers, the factorial is always nonzero. Why is it so?
According to mathguy's comment, the factorial function is positive for all the complex numbers. Is this true? If yes, then why is it so?
Why is the factorial function always nonzero
factorialgamma function
Best Answer
The complex variable function $z! = \Gamma(z+1)$ has no zeros anywhere in the complex plane. Why? One way to see this is to use Euler's reflection formula, $$ z!(-z)! = \frac{\pi z}{\sin \pi z}, \quad z \neq 0 $$
See ProofWiki's "Euler's Reflection Formula" entry for a proof.
If $z!$ were ever zero, then the above formula would also be zero, but that obviously is not the case. In fact $1/(z!)$ is an entire function, that is it is analytic for all $z$, and in particular has no poles, implying $z!$ has no zeros.
EDIT:
This does not necessarily preclude the case where $z!$ has a pole (is infinite) and $(-z)!$ is zero. However, the only poles of $z!$ are at the negative integers, $z=-1, -2, \cdots$, which can be seen from the product formula, $$ z! = \Gamma(z+1) = \Pi_{n=1}^\infty \frac{n}{n+z}\Big( 1+\frac{1}{n}\Big)^z $$ convergent for all $z \neq -1, -2, \cdots$. At these values $-z$ is a positive integers and then $(-z)!$ is known to be non zero.