Q:

Serre's proof that zeta function is meromorphic

I try to understand the proof of Chap. VI, n° 3.1, Prop. 10 in Serre's "A course in arithmetic" (page 70). The goal is to prove that zeta-function can be written as
\begin{align*}
\zeta(s)=\frac{1}{s-1}+\phi(s)
\end{align*}
with an holomorphic $\phi$ for any $s\in\mathbb{C}$ with $\mathfrak{Re}(s)>0$.
I understand the proof except for one detail: $\phi$ is given explicitly by $\phi=\sum_{n=1}^\infty \phi_n$ with
\begin{align*}
\phi_n(s)=\int_n^{n+1} \left(n^{-s}-t^{-s}\right) dt.
\end{align*}
To prove the convergence of $\phi$, we want to show that 
\begin{align*}
\left|\phi_n(s)\right|\leq\frac{|s|}{n^{x+1}} \quad \text{with}\quad x=\mathfrak{Re}(s)\quad(*)
\end{align*}
For this Serre first notes that
\begin{align*}
\left|\phi_n(s)\right| \leq \sup_{n\leq t\leq n+1}\left| n^{-s}-t^{-s}\right|
\end{align*}
Which is clear since the range of integration is $(n+1)-n=1$. Then he sais that the derivative of $n^{-s}-t^{-s}$ is equal to $\frac{s}{t^{s+1}}$. And from this somehow follows (*). My question is: How exactly does this follow? 
I noted the following statements for $s=x+iy$ where $x=\mathfrak{Re}(s)$ and $y=\mathfrak{Im}(s)$ and $f(t)=n^{-s}-t^{-s}$

Because $|t^{iy}|=|e^{iy\ln(t)}|=1$ (since it is on the unit circle) and $|t^{x+1}|=t^{x+1}$ (since $t$ and $x$ are real) we see that
\begin{align*}
\left|f'(t)\right|=\left|\frac{s}{t^{x+1+iy}}\right|=\frac{|s|}{|t^{x+1}| |t^{iy}|}=\frac{|s|}{t^{x+1}}
\end{align*}
Because $f(n)=0$ we can calculate
\begin{align*}
|f(n+1)|=|f(n+1)-f(n)|=\left|\int_n^{n+1}f'(t)dt\right|\leq\sup_{n\leq t\leq n+1}\left|f'(t)\right|=\sup_{n\leq t\leq n+1}\frac{|s|}{t^{x+1}}=\frac{|s|}{n^{x+1}}
\end{align*}

Am I close? :-)

A:

By the definition of $\phi_n$ and the Fundamental theorem of calculus we have
$$|\phi_n(s)| = \left| \int_n^{n+1} f(t) \, dt \right| \stackrel{f(n)=0}{=} \left| \int_n^{n+1} \left(\int_n^t f'(y) \, dy \right) \, dt \right| \leq \int_n^{n+1} |f'(y)| \, dy$$
Use your calculation from 1. (and note that $y \in [n,n+1]$, hence $y\geq n$).

