Order of growth of $(s-1)\zeta(s)$ Again, order of growth problems.

Show that the function $(s-1)\zeta(s)$ is an entire function of growth order $1$; or equivalently,
  $$|(s-1)\zeta(s)| \leq A_{\epsilon} \; \exp \left(a_{\epsilon}|s|^{1+\epsilon} \right).$$

Of course, $\zeta (\cdot)$ denotes the Riemann zeta function.
Many thanks in advance.

I am not sure which textbook you are using. It is therefore a bit of guessing which theorems we are allowed to use. The proof below works if you are using Stein and Shakarchi's Complex Analysis.
At the bottom of Page 171 of the aforementioned textbook, we have the following equation.
$$\xi(s) = \frac{1}{s-1} - \frac{1}{s} + \int_1^\infty (u^{(-s/2)-1/2} + u^{(s/2)-1}) \psi(u) du $$
where
$$\psi(u) = \sum_{n=1}^\infty e^{-\pi n^2 u}.$$
Let $s=\sigma+it$ and $|\sigma|\leq 2N-2 < |\sigma|+2$. For the integral, we have
$$
\begin{eqnarray}
  &&     \Bigg|\int_1^\infty (u^{(-s/2)-1/2} + u^{(s/2)-1}) \psi(u) du \Bigg|  \\
  &\leq& \int_1^\infty (u^{(|\sigma|/2)-1/2} + u^{(|\sigma|/2)-1}) \sum_{n=1}^\infty e^{-\pi n^2 u} du  \\
  &\leq& 2 \sum_{n=1}^\infty \int_1^\infty u^{(|\sigma|+1)/2-1} e^{-\pi n^2 u} du  \\
  &\leq& 2 \sum_{n=1}^\infty \int_0^\infty u^{(|\sigma|+2)/2-1} e^{-\pi n^2 u} du  \\
  &\leq& 2 \Gamma\left(\frac{|\sigma|}{2}+1\right) \sum_{n=1}^\infty \frac{1}{(\pi n^2)^{|\sigma|/2+1}}  \\
  &\leq& 2 \Gamma\left(\frac{|\sigma|}{2}+1\right) \sum_{n=1}^\infty \frac{1}{\pi n^2}  \\
  &=&    \frac{\pi}{3} \Gamma\left(\frac{|\sigma|}{2}+1\right)  \\
  &<&    \frac{\pi}{3} N! \leq \frac{\pi}{3} N^N = \frac{\pi}{3} e^{N\log N} \leq \frac{\pi}{3} e^{(|\sigma|+2)\log (|\sigma|+2)}.
\end{eqnarray}
$$
Since $e^{|\sigma| \log|\sigma|}$ is of growth order 1, this proves that except for the two simple poles at 0 and 1, $\xi(s)$ is of growth order 1.
Furthermore, we have
$$ (s-1)\zeta(s) = (s-1) \pi^{s/2} \frac{\xi(s)}{\Gamma(s/2)}. $$
Now the two poles at 0 and 1 of $\xi(s)$ are cancelled by $\frac{1}{\Gamma(s/2)}$ and $(s-1)$ respectively. Both $\pi^{s/2}$ and $\frac{1}{\Gamma(s/2)}$ are of growth order 1. Therefore $ (s-1)\zeta(s) $ is also of growth order 1.
(For the order of growth of $\frac{1}{\Gamma(s)}$, refer to Theorem 1.6 in Chapter 6 of Stein and Shakarchi.)
