Define the Gamma function as \begin{align} \Gamma(z):= \frac{1}{G(z)} \quad \forall \, z\in \mathbb{C}\setminus \{0, -1, -2, ... \} \end{align} where \begin{align} G: \mathbb{C} \rightarrow \mathbb{C}, \, \, \, G(z) := \lim_{n\to\infty} G_n(z) \quad \text{and} \quad G_n(z) := \frac{z(z+1) \cdot ... \cdot (z+n)}{n! \cdot n^z} \end{align}
I have to prove that $\Gamma$ is continuous.
To do it, I have to prove that the function $G$ is continuous with zeros exactly at the points $0, -1, -2, ...$
$\textbf{My attempt}:$
First, I have to prove that the sequence $(G_n)_n$ converges: For $z\in\mathbb{C}$ choose $R\in\mathbb{N}$ such that $|z|<R$. So for $N\geq 2R$ we can rewrite \begin{align} G_N(z) = G_{2R-1}(z) \cdot \prod_{n=2R}^{N} \frac{G_n(z)}{G_{n-1}(z)} = G_{2R-1}(z) \cdot \exp \sum_{n=2R}^N \ln \frac{G_n (z)}{G_{n-1}(z)} \end{align} I already proved that the series $S:= \sum_{n=2R}^{\infty} \ln \frac{G_n (z)}{G_{n-1}(z)}$ converges uniformly on $B_R(0)$. Then by continuity of exp, $e^{S_N}$ converges to $e^{S}$ on $B_R(0)$, where $S_N :=\sum_{n=2R}^N \ln \frac{G_n (z)}{G_{n-1}(z)}$. So we can deduce the convergence of $G_N$.
Now, to prove continuity of $G$, I would like to prove that $G_N$ converges uniformly to $G$ on every compact subset $K$ of $\mathbb{C}\setminus \{0, -1, -2, ... \}$. Since $G_N$ continuous, then $G$ is continuous.
I have some problem to prove the uniform convergence. I think I have to use that $e^{S_N}$ converges also uniformly to $e^S$ on $B_R(0)$ , but I don't see how.
Any suggestions? Thanks in advance!