Fourier series: $\lim_{n\to\pm\infty} n^p \hat{f}(n) = 0$

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Let $f:\mathbb{R}\to\mathbb{C}$, $f\in C^\infty$ (differentiable infinitely many times) and periodic,$T=2\pi$. Prove that for every $p>0$: $$ \lim_{n\to\pm\infty} n^p \hat{f}(n) = 0$$

So I know that $\lim_{n\to\pm\infty} \left| \hat{f}(n)\right| = 0$ (= Riemann–Lebesgue lemma), but of course this alone is not enough.

I also tried to look at the limit: $$\lim_{n\to\pm\infty} n^p \frac{1}{2\pi} \int_0^{2\pi} f(t)e^{-int}dt$$

I thought about evaluating the integral somehow. Integrating by parts isn't possible because we don't know $F(x)$ exists.

What am I missing then?

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Let $c_n(f)=\hat{f}(n)$. Then $c_n(f^{(k)})=(in)^k c_n(f)$ for all $k\geqslant 0$ by integration by parts (you can do integration by parts!). Now for $k>p$ (note $f\in C^\infty$) we have $|n^p c_n(f)|=|n^{p-k}c_n(f^{(k)})|\rightarrow 0$ (using the Riemann Lebesgue lemma for $f^{(k)}$).