Fourier transform $\mathcal{F}$ of smooth functions $f\in C^l$ decays fast.

I want to prove the above statement, which motivates why the Fourier transform on the Schwartz space is automorphic.

Let $f\in C^l(\mathbb{R}^d)$ and $\alpha$ a multiindex of at most order $l$. From the definition of the Fourier transform we know$$\mathcal{F}[D^\alpha f] =(ik)^{|\alpha|} \mathcal{F}[f]$$So we get

\begin{align}\tilde f(k):=\mathcal{F}[f] &= (ik)^{-|\alpha|} \mathcal{F}[D^\alpha f]\\&= (ik)^{-|\alpha|} (2\pi)^{-d}\int_{\mathbb{R}^d} D^\alpha f e^{-ik\cdot x}dx \\&\leq (ik)^{-l} (2\pi)^{-d} \Vert D^\alpha f e^{-ik\cdot x}\Vert_\infty\end{align}

Conclusion: for every polynomial of order $l$, denoted as $p_l(k)$, we have$$\tilde f(k) \cdot p_l(k) \to 0,\text{ for }k\to \infty, $$So that $\tilde f$ decays faster than any polynomial of order $l$?

Is this proof correct?

The reverse ("fast decaying function has smooth Fourier transform") is then straight forward.