Assume $f$ is periodic over $[0,2\pi]$ and infinitely differentiable.
Then:
$$ \eqalign{ \hat {f' }(r)
&={1\over 2\pi}\int_0^{2\pi} f' (t)\exp(-irt)\,dt\cr
&= {1\over 2\pi} f (t){\exp(-irt) }\Bigl|_0^{2\pi} -{1\over 2\pi} \int_0^{2\pi}(-ir) f (t) {\exp(-irt) }\,dt\cr
&=0+ { ir\over 2\pi} \int_0^{2\pi} f (t) {\exp(-irt) }\,dt \cr
&= { ir} \hat{f }(r).
}
$$
So
$$
\tag{1}ir\hat{f }(r) =\hat {f'}(r).
$$
Applying (1) with $\hat {f'}(r)$ on the left hand side:
$$
ir\hat{f'}(r) =\hat {f''}(r);
$$
whence
$$
-r^2\hat{f }(r)=\hat {f''}(r).$$
Successive iterations yield:
$$
(ir)^n \hat {f }(r) =\widehat {f^{(n)}}(r) .
$$
Since the Riemann-Lebesgue Theorem implies $\widehat {f^{(n)}}(r)$ tends to 0 as $r$ tends to infinity, we have that $r^n \hat f(r) $ tends to 0 as $r$ tends to infinity.