Verifying that $x^i e^{-a|x|^2}$ belongs to the Schwartz space

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According to wikipedia, $x^i e^{-a|x|^2}$ where $i$ is a multi-index, belongs to the Schwartz space. This means that

$$\sup_{x\in\mathbb{R}^n} x^{\alpha} D^{\beta} f(x) < \infty$$

where $f(x) = x^i e^{-a|x|^2}$

It's hard for me to see it because the derivative of that thing is something huge.

First let's break the multiindexes:

$f(x) = x_1^{i_1}x_2^{i_2}\cdots x_n^{i_n} e^{-a|x|^2}$

then $$\partial^\beta f(x) = \frac{\partial^{\beta_1} }{\partial x_1} \frac{\partial^{\beta_2} }{\partial x_2}\cdots \frac{\partial^{\beta_n} }{\partial x_n}x_1^{i_1}x_2^{i_2}\cdots x_n^{i_n} e^{-a|x|^2}$$

Let's first try to see $$\frac{\partial^{\beta_j}}{\partial x_j}f(x) = \frac{\partial^{\beta_j}}{\partial x_j} x_1^{i_1}x_2^{i_2}\cdots x_j^{i_j}\cdots x_n^{i_n} e^{-a|x|^2} = $$

Now things get huge, I need to use the product rule on lots of things. Can somebody give me a help?