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Let $p$ be a rational prime, and let $x$ be an element of $\mathbb{Q}_p$ with $|1-x|_p < 1$. I want to show that $|1-x^{p^n}|_p \leq p^{-p^n}$ for all positive integers $n$, but I'm having a hard time. My idea was to write $$ (1-x)^{p^n} = \sum_{k=0}^{p^n} (-1)^k \binom{p^n}{k}x^k = 1-x^{p^n} + \sum_{k=1}^{p^n - 1} (-1)^k \binom{p^n}{k}x^k $$ so that $$ |1-x^{p^n}|_p \leq \max\left\{|(1-x)^{p^n}|_p, \left|\sum_{k=1}^{p^n - 1} (-1)^k \binom{p^n}{k}x^k \right|_p \right\} = \max\left\{ p^{-p^n}, \left|\sum_{k=1}^{p^n - 1} (-1)^k \binom{p^n}{k}x^k \right|_p \right\}. $$ But I'm stuck there. Can someone please help me out?

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    $p^{-p^n}$ decreases too fast. I don't think this is true even for $n = 1$. For example, take $x = 4$, $p = 3$, $n = 1$. Then $|-63|_3 = 1/9$, but $3^{-3} = 1/27$.2011-08-05
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    On the other hand, you can prove that $|1-x| \leq 1$ implies $|1 - x^{p^n}| \leq p^{-n}$. To show this, use induction and show that $|1-x| \leq p^{-k}$ implies $|1-x^p| \leq p^{-(k+1)}$. This can be easily proven by letting $y = 1-x$, express $1-x^p$ in terms of $y$, and ultrametric inequality2011-08-05

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