Verification of the proof that Gamma functions has no zero https://proofwiki.org/wiki/Zeroes_of_Gamma_Function
Note that $\Gamma(z)= \frac{1}{z} \prod_{n=1}^\infty \frac{(1+1/n)^z}{1+z/n}$. Suppose $\Gamma(z)=0$ for some $z$ in the domain.
The proof in the link says that since each term of the product tends to $1$, the infinite product cannot be zero. However, I think this is a false proof. As far as I know, there are examples such that the terms tend to $1$ but the infinite product might be zero. Am I right?
Moreover, is there a way to prove that the Gamma function does not have a zero instead of using the equation $\Gamma(z)\Gamma(1-z)=\frac{1}{\sin \pi z}$?
 A: The proof given is indeed flawed. However, we can fix the argument with a bit of work.

For a given $z$,
$$
\begin{align}
\log\left(\frac{\left(1+\frac1k\right)^z}{1+\frac zk}\right)
&=z\log\left(1+\frac1k\right)-\log\left(1+\frac zk\right)\\
&=z\left(\frac1k+O\left(\frac1{k^2}\right)\right)-\left(\frac zk+O\left(\frac1{k^2}\right)\right)\\
&=O\left(\frac1{k^2}\right)
\end{align}
$$
Therefore, since $\sum\limits_{k=1}^\infty\frac1{k^2}$ converges,
$$
\log\left(\prod_{k=1}^\infty\frac{\left(1+\frac1k\right)^z}{1+\frac zk}\right)
=\sum_{k=1}^\infty\log\left(\frac{\left(1+\frac1k\right)^z}{1+\frac zk}\right)
$$
converges absolutely.
This means that the product cannot be $0$ unless one of the terms is equal to $0$.

We can verify the formula given for $\Gamma(z)$.
$$
\begin{align}
\prod_{k=1}^n\frac{\color{#C00000}{\left(1+\frac1k\right)^z}}{\color{#00A000}{1+\frac zk}}
&=\color{#C00000}{(n+1)^z}\color{#00A000}{\frac{\Gamma(1+z)\,\Gamma(n+1)}{\Gamma(n+z+1)}}\\
&=\frac{(n+1)^z}{n+z}\frac{\Gamma(n+1)}{\Gamma(n+z)}\Gamma(1+z)
\end{align}
$$
Using Gautschi's Inequality, we get
$$
\frac{(n+1)^z}{n+z}n^{1-z}
\le\frac{(n+1)^z}{n+z}\frac{\Gamma(n+1)}{\Gamma(n+z)}
\le\frac{(n+1)^z}{n+z}(n+1)^{1-z}
$$
Therefore, by the Squeeze Theorem,
$$
\lim_{n\to\infty}\frac{(n+1)^z}{n+z}\frac{\Gamma(n+1)}{\Gamma(n+z)}=1
$$
Thus,
$$
\bbox[5px,border:2px solid #C0A000]{\frac1z\prod_{k=1}^\infty\frac{\left(1+\frac1k\right)^z}{1+\frac zk}=\Gamma(z)}
$$
