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any $x$ which is algebraic of degree $n$ over $\mathbb Q_2$ has a unique series expansion $$ x = \sum 2^{u_j} $$ where $u_j \to \infty$ (unless it is a finite sum)

Then, I would like to prove all rational numbers $u_j$ are rationals with denominator that divides $n!$.

$x$ is root of n-dimmensional polynomial over $\mathbb Q_2$, so $a_nx^n＋a_{n-1}x^{n-1}＋・・・＋a_1x＋a_0＝0$. I tried to see order 2 on both side, but in vain, and I cannot proceed from here. Thank you in advance.

P.S This statement is from Is the algebraic closure of a $p$-adic field complete 's GEdgar's answer. Could you give explanation why denominator decides $n!$?

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    I have some doubts - you're suggesting that any degree $n$ extension of $\mathbb Q_2$ is contained in $\mathbb Q_2(2^{1/n!})$; but $\mathbb Q_2(2^{1/n!})$ is totally ramified over $\mathbb Q_2$, so it will not contain unramified extensions. And it doesn't matter whether you use $n!$ or some bigger integer. But I may be missing something - where does this problem come from? – user8268 Oct 11 '21 at 15:45
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    If, as I assume, $u_j$ is supposed to be a sequence of elements of $\mathbb Q$, then it is not even clear to me how e.g. $2^{1/k}$ is defined in order to make something unique to begin with. After all, there might be $k$ different $k$-th roots of $2$ in our field. – Torsten Schoeneberg Oct 11 '21 at 15:46

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