Equivalence definition of rapidly decreasing in $\mathbb R^n$ I am trying to prove:

a continuous function $f:\mathbb R^n\to \mathbb R$ is rapidly decreasing iff $\sup_{x\in \mathbb R^n}|x|^k|f(x)| <\infty $ for all $k$

where a conitnuous $f$ is defined to be rapidly decreasing iff $\sup_{x\in\mathbb R^n}|x^\alpha f|<\infty$ for all multi-index $\alpha\in\mathbb N^n$. Here $x^\alpha:=x_1^{\alpha_1}x_2^{\alpha_2}...x_n^{\alpha_n}$.
I was trying to show some inequality relations between $|x^\alpha|$ and $|x|^k$. I found that if $1\le |x_i|$ for all $i$, then $|x^\alpha|\le|x|^{2k||\alpha||_{\infty}}$, as by expanding $|x|^{2k}$, we have $x_1^2...x_n^2\le |x|^{2k}$, and therefore $|x^\alpha|\le x_1^{2||\alpha||_\infty}...x_n^{2||\alpha||_\infty}$ as $1\le |x_i|$.
I am stuck here. Any help or comment will be appreciated.

For your first inequality, you actually don't need any case distinction, because for any $i\in\{1,\dots, n\}$ we have $|x_i|\leq \|x\|_2$, so
\begin{align}
|x^{\alpha}|&:=|x_1^{\alpha_1}\cdots x_n^{\alpha_n}|\leq \|x\|_2^{\alpha_1+\dots +\alpha_n}.
\end{align}
For the other, note that clearly, $|x_i|\leq |x_1|+\dots +|x_n|$ for all $i$, so
\begin{align}
\|x\|_2&:=\sqrt{x_1^2+\dots +x_n^2} \leq \sqrt{n}(|x_1|+\dots + |x_n|)
\end{align}
So, if you raise both sides to the power $k$, then on the RHS, by expanding everything out, you get a finite sum of terms of the form $|x^{\alpha}|$; i.e there is a $C>0$ (which is like the maximum of all the combinatorical factors obtained when expanding the RHS) and a finite set $A\subset \Bbb{N}_0^n$ of multindices such that
\begin{align}
\|x\|_2\leq C\sum_{\alpha\in A}|x^{\alpha}|.
\end{align}
