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.