There is a famous theorem says about Gamma function:
Bohr-Mollerup Theorem
Let $f:(0,\infty)\rightarrow \mathbb{R}^+$ be a function satisfying below
(i) $f(x+1)=xf(x), \forall x\in (0,\infty)$
(ii) $f$ is a log-convex function
(iii) $f(1)=1$
Then, $f=\Gamma$ on its domain
This theorem only shows that the Gamma function deserves to be called the factorial function for $x\in\mathbb{R}^+$.
Is it possible to extend this idea to complex plane so Gamma function deserves to be the best factorial function?
Since $t^{z-1}e^{-t}$ can be decomposed into its real part and imaginary part using cosine&sine function, i guess it would show that the Gamma function on complex plane has some flexible property similar to that on real line.
As a function of complex variable, $\Gamma$ is meromorphic. There is an identity theorem: if two meromorphic functions on $\mathbb C$ agree on a set with a limit point in $\mathbb C$, then they agree everywhere in $\mathbb C$. In particular, two meromorphic functions that agree on $(0,\infty)$ agree everywhere. As a consequence, any modification of $\Gamma$ away on $\mathbb C\setminus (0,\infty)$ would lose the property of being meromorphic.
One can use the identity theorem to show that (i) continues to hold in the complex plane. Indeed, $\Gamma(z+1)$ and $z\Gamma(z)$ are two meromorphic functions that agree on $(0,\infty)$; as discussed above, this implies $\Gamma(z+1)=z\Gamma(z)$ for all $z\in\mathbb C$. (Minor detail: at $z=0$ one should interpret the product on the right as the limit $\lim_{z\to 0}z\Gamma(z) $.)