 Solution

If $p=0$ as an element of $K$, the stated lower bound is zero and is immediate. Otherwise set ${\rho }_1=|p{|}^{1/(p-1)}$. The supplied formal group expansion has $g(0)=0$, because $[p](0)=0$. For $z\in \mathcal{M}$, the non-Archimedean estimates giveIf $0<|z|<{\rho }_1$, then $|z{|}^p<|p||z|$, so the two terms cannot cancel and $|[p](z)|=|p||z|>0$. The same strict inequality persists under iteration, so no such $z$ can have finite $p$-power order. Every nonzero $p$-power torsion point therefore satisfies $|z|\ge {\rho }_1$.

For a point of order $p$, $[p](z)=0$ itself forces $|p||z|\le |z{|}^p$, proving the bound for $n=1$. For $n\ge 2$, put $w=[p](z)$, of order $p^{n-1}$. Since $|z|\ge {\rho }_1$, the preceding maximum is $|z{|}^p$. By induction,This proves the formal torsion radius for prime power order. Taking the positive $p$th root gives