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For prime-power contraction in a formal group, a one-dimensional formal group over $R$ has $F(X,Y)=X+Y+$ terms of total degree at least two with coefficients in $R$. Induction under composition therefore givesFor $|x|<1$ this converges and the non-Archimedean inequality gives $|[p](x)|\le \max (|p||x|,|x{|}^2)$. The residue characteristic implies $|p|<1$; in equal characteristic $p$ the linear coefficient is zero, which only improves the bound.

Let $x_n=[p^n](x)$, so $x_{n+1}=[p](x_n)$. If $x=0$ there is nothing to prove. Otherwise put $\rho =\max (|p|,|x|)<1$. Inductively $|x_n|\le |x|$, and then $|x_{n+1}|\le \rho |x_n|$. ThusCompleteness ensures convergence of the formal-series evaluations. The contraction argument in fact does not need the extra discreteness assumption.