[Notation. $a_n \lesssim b_n$ means: there exists some positive constant $C$ s.t. $a_n \le C b_n$.]
A rough'n'ready argument would be:
let
$c_n=\frac{1}{2 \pi} \int_{-\pi}^{\pi}f(y)e^{-i n y}\, dy$
and write
$f(x)=\sum_{n\in \mathbb{Z}} c_n e^{i n x}\quad (1)$
The decay condition on $c_n$ implies uniform convergence of this series:
$\lvert c_n e^{i n x} \rvert \le \lvert n\rvert^{-k}\lvert n^k c_n \rvert \lesssim \lvert n\rvert^{-k}$
and $\sum_{n \in \mathbb{Z}} \lvert n\rvert^{-k}$ is a convergent numerical series. Now differentiate (1) termwise: you get
$\sum_{n \in \mathbb{Z}}i n c_n e^{i n x}$
which is again a uniformly convergent series:
$\lvert i n c_n \rvert \lesssim \lvert n\rvert^{1-k}.$
So (1) is a uniformly convergent series whose term-by-term derivative is uniformly convergent. This implies that $f$ is differentiable and
f'(x)=\sum_{n \in \mathbb{Z}}i n c_n e^{i n x}
so that, again by uniform convergence, f' is also continuous.