Why is it so obvious that $x^k e^{-\frac{x^2}{2}}$ is a Schwartz-function? (Verification)

Question

I wanna show $h_k(x)=x^k e^{-\frac{x^2}{2}}$ , for all $k\ge 0$ , and $k \in \mathbb{N}$ that is Schwartz function.

Is it enough if I prove the following:

Differentiation: I've got that $h_k'(x) = e^{-\frac{x^2}{2}}(kx^{k-1}-x) $ with $\|x\| \rightarrow \infty$ we have that $h_k(x) \rightarrow 0 $ since it's exponential functions decays faster than potential functions.

Thanks!

Answer 1

$h_k$ and any derivative of $h_k$ is a polynomial times $e^{-x^2/2}$, and for any polynomial $p(x)$ we have $\lim_{|x|\to +\infty}p(x) e^{-x^2/2}=0$, so $h_k\in\mathscr{S}(\mathbb{R})$ is trivial.