I am aware of the general property of the Fourier transform that it forms an automorphism of the Schwartz space. However, prior to saying "It forms an automorphism", my Fourier-analysis material states that
The fact that if $f\in \mathscr{S}(\mathbb{R})$ then $\hat{f}\in \mathscr{S}(\mathbb{R})$ follows trivially from the polynomial-derivative dualism
where the polynomial-derivative dualism refers to the fact that $\hat{s'}(v) = 2\pi iv \hat{s}(v), \hat{s}'(v) = -2\pi\hat{ts}(v)$. My material defines the Schwartz space $\mathscr{S}(\mathbb{R})$ as the set of functions $f$ such that $\lim_{|t|\to\infty}t^n\frac{d^m}{dt^m}(f(t)) = 0, \forall n,m\in \mathbb{N}_0$. Hence I am curious how you could argue about the inclusion of $f$'s Fourier transform. Namely, suppose that $f\in \mathscr{S}(\mathbb{R})$ and let $n , m \in \mathbb{N}_0$ be arbitrary. Then $v^n\frac{d^m}{dv^m}(\hat{f}(v)) = v^n(-2\pi i)^m(\hat{t^mf}(v)) = v^n(-2\pi i)^m\int_{\mathbb{R}}t^mf(t)e^{-2\pi i t v}dt$ and I am not sure how to proceed from here as I seem to not have any tools to disassemble the integral in a reasonable way.