How to solve this product? I was reading the Wolfram article on infinite products again, and the product 
$$\prod_{n=1}^{\infty} (1 + \frac{1}{n^k})$$
Is solved on there. I was wondering if anyone knows how to do that? I tried checking the source but couldn't actually find it.
 A: Let $Z = \{ \omega \in \mathbb{C} : \omega^k = -1\}$. Then
$$ 1 + \frac{1}{n^k} = \prod_{\omega \in Z} \left(1 - \frac{\omega}{n} \right). $$
Now plugging this back, the partial product is
$$ \prod_{n=1}^{N} \left( 1 + \frac{1}{n^k} \right)
= \prod_{\omega \in Z} \prod_{n=1}^{N} \left(1 - \frac{\omega}{n} \right)
= \prod_{\omega \in Z}  \frac{\Gamma(N+1-\omega)}{\Gamma(1-\omega)N!}. $$
In view of the Stirling's approximation, we know that
$$ \frac{\Gamma(N+1-\omega)}{N!} \sim \frac{1}{N^{\omega}} \quad \text{as} \quad N \to \infty, $$
and since $\sum_{\omega \in Z} \omega = 0$ for $k \geq 2$, we obtain
$$ \prod_{n=1}^{\infty} \left( 1 + \frac{1}{n^k} \right)
= \prod_{\omega \in Z}  \frac{1}{\Gamma(1-\omega)}. \tag{*} $$
When $k$ is even, then we know that $\omega \in Z$ if and only if $-\omega \in Z$, and in this case, we can group the factor according to this parity and then utilize the Euler's reflection formula to write
$$ \frac{1}{\Gamma(1-\omega)\Gamma(1+\omega)} = \frac{1}{\omega \Gamma(1-\omega)\Gamma(\omega)} = \frac{\sin(\pi \omega)}{\pi \omega}. $$
You may plug this back to simplify the product $\text{(*)}$.
