In the region $\sigma \geq 2$ the $\zeta$-function is bounded:
$$
|\zeta(s)|=\left|\sum n^{-s}\right| \leq \sum n^{-\sigma} \leq \zeta(2)
$$
The same argument shows that also the derivatives of $\zeta$ are bounded in this region, since one can derive termwise the defining $\zeta$-series. So we can restrict our further considerations to the vertical strip $1<\sigma \leq 2$. Then it is enough to show that
$$
\left|\zeta^{(m)}(s)\right| \leq C_{m}|s| \quad(1<\sigma \leq 2,|t| \geq 1) .
$$
For this, we use the integral representation. (It is also possible to use the integral representation
By the product formula, the higher derivatives of $s F(s)$ are a linear combination of $F^{(\nu)}(s)$ and $s F^{(\mu)}(s)$. So it is enough to show that any of the higher derivatives of $F$ is bounded in the vertical strip $1<\sigma \leq 2$. We have:
$$
F^{(m)}(s)=\int_{1}^{\infty}(-\log t)^{m} t^{-s-1} \beta(t) d t .
$$
Using an estimate of the shape
$$
|\log (t)| \leq C_{m}^{\prime} t^{\frac{1}{2 m}} \quad(|t| \geq 1) \quad \text { with a suitable constant } C_{m}^{\prime},
$$
in connection with $|\beta(t)| \leq 1$, we obtain
$$
\left|F^{(m)}(s)\right| \leq C_{m}^{\prime} \int_{1}^{\infty} t^{-\frac{3}{2}} d t<\infty
$$
MATH4079 COURSE NOTES ：
$$
Z(s):=e^{D(s)}
$$
then we have
$$
|Z(\sigma+\mathrm{it})|^{4}|Z(\sigma+2 \mathrm{i} t)||Z(\sigma)|^{3} \geq 1 .
$$
We want to show that this Lemma can be applied to $\zeta(s)=Z(s)$. For this we consider
$$
b_{n}:= \begin{cases}1 / \nu & \text { if } n=p^{\nu}, p \text { prime } \ 0 & \text { else }\end{cases}
$$
Then it holds
$$
D(s)=\sum_{p} \sum_{\nu} \frac{1}{\nu} p^{-\nu s}=\sum_{p}-\log \left(1-p^{-s}\right)
$$
and because of this
$$
e^{D(s)}=\prod_{p}\left(1-p^{-s}\right)^{-1}=\zeta(s) .
$$
We obtain after a trivial transformation