This question arises from the proof of theorem 8 (Weierstrass division) in chapter 2 of Siegfried Bosch's Lectures on formal and rigid geometry. Given the Tate algebra $T_n =K \langle \zeta_1, \dots, \zeta_n \rangle $ over a complete (nonarchimedean) valued field $K$, and a $\zeta_n$-distinguished element $g \in T_n$ of order $s$, for any $f \in T_n$ there exists a unique $q \in T_n$ and unique polynomial $r \in T_{n-1}[ \zeta_n ]$ of degree $< s$ in $\zeta_n$ such that $f=gq+r$.
Bosch seems to state that this follows from the fact $\vert f \vert = \max ( \vert g \vert \vert q \vert, \vert r \vert) $ but I can't see how.
In the case where $\vert g \vert = 1$ (this is a valid assumption for proving the theorem), $g$ being distinguished of order $s$ means, upon reduction to the residue field $k$, where $\tilde{g}$ is the reduction of $g$, that $\tilde{g} = a_s \zeta_n ^s + \dots + a_0 \zeta_n ^0$, where $a_s$ is in $k^{\ast}$ and $a_i \in k[\zeta_1, \dots, \zeta_{n-1}]$. That $a_i$ is a polynomial and not a power series follows from the fact the Tate algebra consists of power series whose coefficients go to $0$, so all but finitely many coefficients lie in the maximal ideal of the valuation ring of $K$.
If the "order" is what I think it is and behaves how it should behave (I leave that to OP to check), the following should work:
Let $f=gq_1+r_1 = gq_2+r_2$ i.e. $$g(q_2-q_1)=r_2-r_1.$$
The right hand side has degree and hence order $< s$, but $g$ is of order $s$ by assumption, so the left hand side has order $\ge s$ unless $q_2=q_1$. That of course implies $r_1=r_2$.
Reference for the one-dimensional case: P. Schneider, Theorie des Anstiegs, Satz 9.11. (Clash of notation: Schneider's $P$ is Bosch's $g$, Schneider's $G$ is Bosch's $q$, Schneider's $Q$ is Bosch's $r$.)