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.

Another approach is through the Laurent series of the Riemann zeta function at $s=1$,
$$\zeta(s)=\frac1{s-1}+\sum_{n=0}^\infty\frac{(-1)^n}{n!}\gamma_n(s-1)^n\;,$$
where the $\gamma_n$ are the Stieltjes constants. Multiplying by $s-1$ yields
$$(s-1)\zeta(s)=1+\sum_{n=0}^\infty\frac{(-1)^n}{n!}\gamma_n(s-1)^{n+1}\;.$$
Theorem 2.2.2 of Entire Functions by Ralph Philip Boas expresses the order $\mu$ of an entire function given by a power series
$$f(z)=\sum_{n=0}^\infty a_nz^n$$
in terms of the coefficients:
$$\mu=\limsup_{n\to\infty}\frac{n\log n}{\log (1/|a_n|)}\;.$$
To evaluate the limit superior, we can use bounds found by Matsuoka: For all $n\ge10$,
$$|\gamma_n|\le\frac{\exp(n\log\log n)}{10000}\;,$$
and for infinitely many n
$$|\gamma_n|\gt\exp(n\log\log n-n\epsilon)\;.$$
By substituting Stirling's approximation for the factorial,
$$\log n!\sim n\log n -n\;,$$
we can see that the limit superior is $1$: The upper bound on $\gamma_n$ ensures that the quotient is eventually below $1+\delta$, and the lower bound ensures that it is infinitely often above $1$.
