(The other posts on this site with similar topic don't seem to answer this very clearly).
Let $f,g \in \mathcal{S}(\mathbb{R})$ so that for, i.e., $f$:
$$ \lim_{|x| \to \infty} |x|^k |f^{(l)}(x)| = 0 \text{ for all integers $k$, $l \ge 0$}. $$
Then why does $fg \in \mathcal{S}(\mathbb{R})$?
Attempt: The product rule yields $fg = f'g + fg'$, but it's not immediately clear how this leads to a solution.
Leibniz rule: $(fg)^{l}(x)=\displaystyle\sum_{m=0}^{l}(l,m)f^{(m)}(x)g^{(l-m)}(x)$ and each $\sup|x^{k}f^{(m)}(x)|<\infty$ and also $\sup|g^{(l-m)}(x)|<\infty$.
Actually one can show that the definition is equivalent to the following one: \begin{align*} \sup_{x\in{\bf{R}}}|x^{k}f^{(l)}(x)|<\infty. \end{align*}