[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.
Hint: Use the theorem
If $g_n \to g$ pointwise, and $g_n$ differentiable, and that $g'_n \to h$ uniformly, then $g$ is differentiable with derivative $h$.
Let $g_n$ in this case be the Fourier partial sums $\sum_{|k| \leq n} a_k e^{ikx}$.
Best Answer
Yes. Smoothness is equivalent to the Fourier coefficients forming a sequence that decays rapidly (faster than any polynomial). To see the direction you asked about, note that if $\{c_n\}$ is a rapidly decaying sequence, then the sum $\sum c_n e^{in x}$ will converge uniformly as will every derivative. So the sum will represent a smooth function. (The continuous analog of this is that the Fourier transform induces an isomorphism of the Schwartz space onto itself.)
In general, one can use the Fourier coefficients to give an $L^2$ definition of differentiability. Morally, a function ought to be $C^k$ if its Fourier coefficients decay like $|n|^{-k}$. This is not quite accurate, but one can use this to define the Sobolev spaces on the circle (more generally on the torus, or using the Fourier transform on euclidean space, and by an extension process on compact manifolds) that describe how many "weak" derivatives a function has. This sort of argument shows that $L^2$ differentiability is roughly comparable with normal differentiability, and using such Sobolev spaces turns out to be integral when one tries to prove explicit bounds about things such as elliptic regularity. There, working with Fourier coefficients (or transforms) and a multiplication operator is significantly easier than working in the original function space with the correspondingly more complicated differential operator. (I am using the fact that constant-coefficient differential operators correspond under the Fourier transform to multiplication by a polynomial.)