Order of growth of $ \prod_{n=1}^{+\infty} (1-e^{-2\pi n}\cdot e^{2\pi i z})$ The order of an entire function $f$ id defined as $$ord ( f) = inf \{\lambda \geq 0  \ | \ \exists A, B > 0 \ s.t. \  |f(z)|\leq Ae^{B|z|^{\lambda}} \forall z \in \mathbb{C} \}$$ 
I have $F(z) = \prod_{n=1}^{+\infty} (1-e^{-2\pi n}\cdot e^{2\pi i z})$. 
I want to prove that $F$ is entire of order at most $2$ but I'm stuck on it,any hint ?
 A: You can bound the modulus of the product by bounding the modulus of each factor.
$$\left\lvert 1 - e^{-2\pi n}\cdot e^{2\pi iz} \right\rvert \leqslant 1 + e^{-2\pi n} \left\lvert e^{2\pi iz}\right\rvert = 1 + e^{-2\pi(\operatorname{Im} z + n)} \leqslant 1 + e^{2\pi(\lvert z\rvert - n)}.$$
Split the product at $n \approx \lvert z\rvert$ and estimate both parts separately to get constants $A,B$ satisfying
$$\left\lvert F(z)\right\rvert \leqslant A\cdot e^{B\cdot \lvert z\rvert^2}.$$
Namely, splitting at $k(z) := \lceil \lvert z\rvert \rceil$ (for $z \neq 0$), the second part can be estimated
$$\begin{align}
\prod_{n=k(z)}^\infty \left( 1 + e^{2\pi(\lvert z\rvert -n)}\right)
&= \exp \left(\sum_{n=k(z)}^\infty \log \left( 1 + e^{2\pi(\lvert z\rvert -n)}\right)\right)\\
&< \exp \left(\sum_{n=k(z)}^\infty e^{2\pi(\lvert z\rvert - n)}\right)\\
&\leqslant \exp \left(\sum_{m=0}^\infty e^{-2\pi m}\right)\\
&= \exp \left(\frac{e^{2\pi}}{e^{2\pi}-1}\right),
\end{align}$$
and the first part
$$\begin{align}
\prod_{n=1}^{k(z)-1} \left( 1 + e^{2\pi(\lvert z\rvert -n)}\right)
&< 2^{k(z)-1}\prod_{n=1}^{k(z)-1} e^{2\pi(\lvert z\rvert - n)}\\
&< 2^{k(z)-1} \prod_{n=1}^{k(z)-1} e^{2\pi \lvert z\rvert}\\
&= \exp \left((k(z)-1)\left(2\pi \lvert z\rvert + \log 2\right)\right)\\
&< \exp \left(\lvert z\rvert(2\pi\lvert z\rvert + \log 2)\right)\\
&< 2e^{(2\pi+\log 2)\lvert z\rvert^2}\\
&< 2e^{7\lvert z\rvert^2}.
\end{align}$$
Putting the two parts together, we have (also for $z = 0$)
$$\lvert F(z)\rvert < 6 e^{7\lvert z\rvert^2},$$
showing the order of $F$ is at most $2$. On the other hand, choosing $z_k = \frac12 - ki,\, k \in \mathbb{Z}^+$, we see
$$\lvert F(z_k)\rvert \geqslant \prod_{n=1}^k \left(1 + e^{2\pi (k-n)}\right) > \prod_{n=1}^k e^{2\pi(k-n)} = e^{\pi k(k-1)},$$
so the order of $F$ is at least $2$, hence exactly $2$.
