Let $f:B\to\mathbb{C}$ a holomorphic function in the horizontal strip $B:=\{z\in \mathbb{C} : -1 < \mathrm{Im}\:{z} < 1\}$. Assume that we have $\beta > 0$ such that for every $z \in B$, it is $|f(z)| \leq \left (1+|z|\right )^\beta$. We want to prove that, for every $n\in\mathbb{N}$, we have the following inequality $$ |f^{(n)}(x)| \leq n!\, 2^\beta \left( 1+|x| \right)^\beta,\: x \in \mathbb{R} $$ . Our first approach to the problem has been to try to use Cauchy's inequalities to bound the value of the function in the real line, but we couldn't find a way to reach the asked inequality using closed disks contained in $B$.
Then, another approach we considered was to transform continuoulsy the band $B$ into the unit disk $\mathbb{D}$, using the transformation $\zeta :B\to\mathbb{D}$, defined by $\zeta(z) = \frac{e^\frac{\pi z}{2} -1}{e^\frac{\pi z}{2} +1}$. Then, the inverse map $\zeta^{-1}(w) = \frac{2}{\pi}\log{\frac{w+1}{w-1}}$ transforms $\mathbb{D}$ into $B$ continuously. If this is correct, and assuming that $\zeta$ maps the real axis to the boundary of a closed disk living in $\mathbb{D}$, it seems that we could reach a condition where the hypothesis of Cauchy's inequalities hold, and maybe calculate the asked bound. However, we don't know if this approach has any flaw in its reasoning or even if the map $\zeta$ has the geometric properties needed. Thanks in advance for your help.
Your first idea works. For $x \in \Bbb R$ and $0 < r < 1$ one can apply Cauchy's integral formula to the closed disk centered at $x$ and radius $r$. For $|z-x| = r$ is $$ |z| \le r+|x| \implies |f(z)| \le (1+r+|x|)^\beta $$ so that $$ |f^{(n)}(x)| \le n! \frac{\max \{ |f(z)| : |z-x| = r \} }{r^n} \le n!\frac{(1+r+|x|)^\beta}{r^n} \, . $$ With $r \to 1$ it follows that $$ |f^{(n)}(x)| \le n! (2+|x|)^\beta \le 2^\beta(1+|x|)^\beta \, . $$