Let $U:=\{s \in \mathbb{C} | \operatorname{Re}(s)>0\} .$ I know that $$ \Gamma(s):=\int_{0}^{\infty} e^{-x} x^{s-1} d x \quad\left(x^{s}=e^{s \log x}\right) $$ defines a holomorphic function $\Gamma: U \rightarrow \mathbb{C}$ that satisfies the functional equation $s \Gamma(s)=\Gamma(s+1) .$ I've already constructed a holomorphic extension of $\Gamma$ to $\tilde{U}:=\mathbb{C} \backslash\{0,-1,-2, \ldots\}$:
$\Gamma(s)=\frac{1}{s(s+1)(s+2)...(s+n-1)} \Gamma(s+n)$
Now I have to argue that this is the ONLY extension of $\Gamma$ from $U$ to $\tilde{U}$ How do I have to argue?
By the Identity theorem (also known as principle of analytic continuation), if two holomorphic functions are equal on a open connected set they are equal on all their common domain. Thus, since $U$ is an open and connected set, every holomorphic function defined on $\tilde{U}$ which extends $\Gamma$ will be equal to the analytic continuation you obtained.