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Let $|\cdot|$ be an ultrametric absolute value (i.e it is a function from a field $\mathcal{K}\rightarrow [0,\infty) $ that satisfies that $|\alpha \beta| = |\alpha||\beta|$ and $|\alpha + \beta| \leq \max \{ |\alpha|, |\beta|\}$, and $|\alpha| = 0$ if and only if $\alpha = 0$).

I want to show that if $\beta$ is a root of the polynomial $$X^n + \alpha_1 X^{n-1} + \cdots + \alpha_n = 0$$ where $|\alpha_i| \leq 1$ for all $i$ ,then $|\beta| \leq 1$.

I've tried several things with the ultrametric inequality like showing that $$|\beta||\beta^{n-1}+ \alpha_1 \beta^{n-2}+ \cdots +\alpha_{n-1}| = |\alpha_n| \leq 1,$$ and I've tried to bound the factor $|\beta^{n-1}+ \alpha_1 \beta^{n-2} + \cdots +\alpha_{n-1}|$ using what we know about the $\alpha_i$ and the ultrametric inequality, but I'm just going in circles. Any hints or solutions would be greatly appreciated.

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Since $-\beta^n = \alpha_1 \beta^{n-1} + \dots + \alpha_n$, we get $|\beta|^n = |\alpha_1 \beta^{n-1} + \dots + \alpha_n|$. Suppose $|\beta| > 1$. Then, $$|\beta|^n = |\alpha_1 \beta^{n-1} + \dots + \alpha_n| \leq \max \{ |\alpha_1||\beta|^{n-1},\dots,|\alpha_n| \} \leq \max \{ |\beta|^{n-1},\dots,1 \} = |\beta|^{n-1}.$$ Hence, $|\beta|^n \leq |\beta|^{n-1} \implies |\beta| \leq 1$, a contradiction.

Thus, $|\beta| \leq 1$.

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