Question about hint from an exercise from Stein's Complex Analysis. Finding uniform constant bounds for infinite products I am having difficulty following the hint of this exercise from Stein and Shakarchi's Complex Analysis. This is Exercise 4 of Chapter 5.
Let $t>0$ be given and fixed, and define $F(z)$ by 
$$F(z) = \Pi_{n=1}^\infty (1-e^{-2\pi nt} e^{2\pi iz}).$$
(a) Show that $|F(z)| \le Ae^{a|z|^2}$. 
The following hint is basically the proof. However, I cannot fill in some details. First, the constants of this bound must be uniform over $z$, so in the first bound of the hint, we should have to be able to choose $N$ independent of $z$ to have $(\sum_{N+1}^\infty e^{-2\pi nt})e^{2\pi |z|} \le 1$. But I can't see how we can get this $N$ that works for all $z$. 
Next, how does this bound over the sums give the bound for the product, i.e. $|F_2(z)| \le A$? 
Finally, for the final bound on $F_1$, how do we get $2^N e^{2 \pi N |z|} \le e^{c' |z|^2}$? 
I would greatly appreciate some help.

 A: The authors write $N\approx c|z|!$ Thus there won't be one $N$ that works for all $z.$
Given $z,$ choose $N$ such that $|z|/t\le N\le |z|/t+1.$ Then
$$|F_2(z)| \le \prod_{n=N+1}^{\infty}(1+e^{-2\pi nt}e^{2\pi |z|})$$ $$ \le \exp \left ( \sum_{N+1}^{\infty}e^{-2\pi nt}e^{2\pi |z|}\right )$$ $$ = \exp \left( e^{-2\pi (N+1)t}e^{2\pi |z|}/(1-e^{-2\pi t})\right)\le \exp (1/(1-e^{-2\pi t})).$$
Thus we have a bound on $|F_2(z)|$ independent of $z.$
Added later For the other estimate, let's start with $|F_1(z)|\le 2^Ne^{2\pi N|z|}.$ Now
$$2^Ne^{2\pi N|z|} < e^Ne^{2\pi N|z|} \le e^{|z|+1}e^{2\pi (|z|+1)|z|},$$
where we have used $N\le |z|+1.$ The expression on the right has the form $e^{a+b|z|+c|z|^2},$ where $a,b,c$ are positive constants. This is not bounded above by $e^{c'|z|^2}$ for any $c'>0.$ The problem is with $|z|$ small, not $|z|$ large (the inequality falls apart when $|z|=0$). So they made a mistake in the hint there, but it's no problem. Remember, the ultimate aim is to show $|F_1(z)|\le Ae^{\alpha |z|^2}$ for some positive $A,\alpha.$ Writing $A=e^\beta,$ we want
$$e^{a+b|z|+c|z|^2}\le e^{\beta +\alpha |z|^2}$$
for some choice of $\alpha,\beta >0.$ That's easy: Let $\alpha = b+c,\beta =a+b+c.$ (Think about $|z|\le 1, |z|\ge 1$ separately.)
