Why does $s(s-1)\xi(s)$, have order of growth 1?
In other words, why is it that $\forall \epsilon > 0 $ $\exists A_{\epsilon},B_{\epsilon} \in \mathbb R_+$ so that $\forall s \in \mathbb C$, $(s)(s-1)\xi(s) \le A_{\epsilon}exp^{B_{\epsilon}|s|^{1+\epsilon}}$
Recall that $\xi(s)$ is defined by the functional equation: $$ s(s-1)\zeta(s) = \frac{s(s-1)*\xi(s)*\pi^{s/2}}{\Gamma(s/2)} \tag{*} $$
$$$$ Note also that if I can figure out why $\xi(s)$ has order of growth 1, then I can quickly determine that $$ |(s-1)\zeta(s)| \leq A_{\epsilon} \; \exp \left(a_{\epsilon}|s|^{1+\epsilon} \right) \tag{**} $$ because I can show that $\frac{1}{\Gamma(s/2)}$, $(s-1)$, and $\pi^{s/2}$ each have of order of growth 1. Hence, I can plug in their growth orders into (*) and deduce (**).