In "Mathematical Methods for Physicists" (Arfken & Weber, 7th ed.), exercise 13.1.16 says the following,
Prove that $$|\Gamma (\alpha+i\beta)|=|\Gamma(\alpha )|\prod_{n=0}^{\infty}\left[1+\frac{\beta^2}{(\alpha+n)^2}\right]^{-1/2}$$
where $\alpha,\beta\in\mathbb{R}$.
However, when I try to show this I come up with an additional factor of $\left[1+\left(\frac{\beta}{\alpha}\right)^2\right]^{-1/2}$. I'm sure the way to do this is through the Weierstrass product formula for the gamma function, which says that
$$\frac{1}{\Gamma (z)}=ze^{\gamma z}\prod_{n=1}^{\infty}\left(1+\frac{z}{n}\right)e^{-\frac{z}{n}}$$
where $\gamma=0.57721\cdots$ is the Euler-Mascheroni constant. My attempt goes as follows:
$$ \begin{align*} \frac{1}{|\Gamma(\alpha+i\beta)|^2}&=\frac{1}{\Gamma (\alpha+i\beta)}\frac{1}{(\Gamma (\alpha+i\beta))^*}\\ &=\frac{1}{\Gamma (\alpha+i\beta)}\frac{1}{\Gamma (\alpha-i\beta)}\\ &=\left[(\alpha+i\beta)e^{\gamma (\alpha+i\beta)}\prod_{n=1}^{\infty}\left(1+\frac{(\alpha+i\beta)}{n}\right)e^{-\frac{\alpha+i\beta}{n}}\right]\times\left[\textrm{same but }i\beta\to-i\beta\right]\\ &=(\alpha^2+\beta^2)e^{2\gamma\alpha}\prod_{n=1}^{\infty}\left[\left(1+\frac{\alpha}{n}\right)^2+\frac{\beta^2}{n^2}\right]e^{-\frac{2\alpha}{n}}\\ &=(\alpha^2+\beta^2)e^{2\gamma\alpha}\prod_{n=1}^{\infty}\left[\left(1+\frac{\alpha}{n}\right)^2\left(1+\frac{\beta^2}{(\alpha+n)^2}\right)\right]e^{-\frac{2\alpha}{n}}\\ &=\left(1+\frac{\beta^2}{\alpha^2}\right)\left\{\alpha^2 e^{2\gamma\alpha}\prod_{n=1}^{\infty}\left(1+\frac{\alpha}{n}\right)^2e^{-\frac{2\alpha}{n}}\right\}\prod_{n=1}^{\infty}\left[1+\frac{\beta^2}{(\alpha+n)^2}\right]\\ &=\left(1+\frac{\beta^2}{\alpha^2}\right)\frac{1}{|\Gamma (\alpha )|^2}\prod_{n=1}^{\infty}\left[1+\frac{\beta^2}{(\alpha+n)^2}\right] \end{align*} $$
Which by taking the inverse and square root on both sides gives me the (incorrect) result,
$$\boxed{|\Gamma(\alpha+i\beta )|=\color{red}{\left(1+\frac{\beta^2}{\alpha^2}\right)^{-1/2}}|\Gamma(\alpha )|\prod_{n=1}^{\infty}\left[1+\frac{\beta^2}{(\alpha+n)^2}\right]^{-1/2}}$$
How do I get get rid of the term which I've colored red (above)? Where did I go wrong?
Not sure if it is your book's problem or what, but the Weierstrass product formula is $$ \frac1{\Gamma(z)} = ze^{\gamma z} \prod_{n=\color{red}1}^\infty \left(1+\frac zn\right)e^{-z/n} $$ (Note that $(1+\frac z0)e^{-z/0}$ is undefined, so the product cannot start from 0.)
So after taking inverse square root your solution should be $$ \left| \Gamma(\alpha + i\beta) \right| = \color{blue}{\left( 1 + \frac{\beta^2}{\alpha^2}\right)^{-1/2}} \left| \Gamma(\alpha) \right| \prod_{n=\color{red}1}^\infty \left( 1 + \frac{\beta^2}{(\alpha+n)^2}\right)^{-1/2} $$ and the extra factor is exactly the $n=0$ term.