I want to show that $\Gamma(z)\Gamma(1-z)=\pi\csc \pi z$ for $z\notin\mathbb{Z}$ by using the infinite product representation of $\Gamma(z)$ and $\sin \pi z$. Also I know that the convergence of the infinite product in both cases is uniform over compacts of $\mathbb{C}\setminus\mathbb{Z_\leq 0}$.
The usual proof I've seen goes like this:
First, using $\Gamma(z+1)=z\Gamma(z)$ we have that $\Gamma(1-z)=-z\Gamma(-z)$
Now $\Gamma(z)=\frac{e^{-\gamma z}}{z}\prod_{n=1}^{\infty} \frac{1}{1+\frac{z}{n}} e^{\frac{z}{n}}$ so $-z\Gamma(z)\Gamma(-z)=\frac{1}{z}\prod_{n=1}^{\infty} \frac{1}{1+\frac{z}{n}}e^{\frac{z}{n}}\prod_{m=1}^{\infty} \frac{1}{1-\frac{z}{m}}e^\frac{-z}{m}=\frac{1}{z}\prod_{n=1}^\infty \frac{1}{(1+\frac{z}{n})(1-\frac{z}{n})}=\pi \csc\pi z$.
My question is, why is correct to put the same index in both products? Is there a way to justify formally? Any proofs or references are welcomed! I've seen this question a lot of times but I didn't find an answer to the product manipulation.
If $\prod_{n=1}^{\infty} a_n$ and $\prod_{n=1}^{\infty} b_n$ are convergent infinite products, with limits $A$ and $B$, then so is $\prod_{n=1}^\infty a_n b_n$ with limit $AB$. Basically this is continuity of multiplication.