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Consider a prime $p$. Let $f$ be the Euler series defined by $f(z)=\sum_{n\ge0}n!z^n\}$. It is defined and analytic over $\mathcal D=\{z\in\mathbb C_p\mid v_p(z)>-\frac1{p-1}\}$. I try to check if this function admits zeroes on $\mathcal D$. I tried to compute the Newton polygon of $f$ in order to apply Proposition 1 of Chapter 5 Section 2 of Bruhat lectures (here: http://www.math.tifr.res.in/~publ/ln/tifr27.pdf). But no luck. I did not manage.

My question: Does $f$ admit zeroes in $\mathcal D$?

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    $\begingroup$ This power series converges on the closed unit disc $\{z : |z|_p \leq 1\}$, so why are you looking only in the disc $\{z : |z|_p < (1/p)^{1/(p-1)}\}$? And a power series in $1 + z\mathbf Z_p[[z]]$ can't have any zeros with $|z|_p < 1$ since all terms have absolute value less than 1 besides the constant term 1. To count zeros in the closed unit disc, use Strassmann's theorem (all zeros in the closed unit disc will be algebraic over $\mathbf Q_p$). $\endgroup$ Commented Aug 14, 2020 at 18:44

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