Prove $|n^s| = n^{Re(s)}$ How can one prove that $|n^s|= n^{Re(s)}$, where $n \in \mathbb{N}$ and $s \in \mathbb{C}$, such that $Re(s) > 1$  (I found this on the link Convergence of $\zeta(s)$ on $\Re(s)> 1$)?
I tried: $|n ^s|= |e^{ s log n}| = | e^{ s( ln(n) + i Arg (n) + 2 k \pi i) } |= |e^{ s( ln(n) + 2 k \pi i) }| = | e^{ (x + iy ) (ln (n) + 2 k \pi i) } | = | e^{x ln (n) + i y ln(n) +2 k \pi x i - 2 k \pi y )} |  = | e^{x ln (n) - 2 k \pi y } | = n^x e^{- 2 k \pi y } $.
How do we obtain $n^x$?
 A: Try using the definition $|z| = \sqrt{\overline{z}z}$, for $s = x+iy$
\begin{align*}
|n^s| & = \sqrt{\overline{n^s}\cdot n^s} \\
& = \sqrt{\overline{n^{x+iy}}\cdot n^{x+iy}} \\
& = \sqrt{n^{x-iy}\cdot n^{x+iy}} \\
& = \sqrt{n^{x}n^{-iy} n^{x}n^{iy}} \\
& = \sqrt{n^{2x}n^{iy-iy}} \\
& = \sqrt{n^{2x}n^{0}} \\
& = \sqrt{n^{2x}} \\
& = n^{x} \\
\end{align*}
A: Assuming that $s = x + iy$, and following this link, then:
$$n^s = n^{x+iy} = n^x \cdot n^{iy} = n^x \cdot e^{iy\log(n)}.$$
Recall that:
$$e^{iy\log(n)} = \cos(y \log(n)) + i \sin(y \log(n)),$$
and hence:
$$|e^{iy\log(n)}| = \sqrt{\cos^2(y \log(n)) + \sin^2(y \log(n))} = 1.$$
As a consequence:
$$|n^s| = |n^x| \cdot |e^{iy \log(n)}| = n^x \cdot 1 = n^x.$$
But $x = \text{Re}(s)$, and hence:
$$|n^s| = n^{\text{Re}(s)}.$$
A: On one hand we could have this reasoning:
$n^s=n^{a+ib}=n^a\times n^{ib}$
And since $n^{ib}=\exp(ib\ln n)=e^{i\theta}\quad$ with $\theta=b\ln n$ so it is of module $1$.
Yet actually the premises $n^{a+ib}=n^a\cdot n^{ib}$ is false in general if you consider complex exponentiation.
But we generally assume the definition $b^z=\exp(z\ln b)$ when $b$ is a positive real, where $\exp$ is defined by its power series.
See this Definition of exponential function, single-valued or multi-valued?
Therefore it is really a matter of conventions, the result $|n^z|=n^a$ is fine according to this latter convention, but yours is the correct one assuming the complex exponentiation convention.
