= 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 give
$$
|f(z)|=|z|,\qquad |g(z^p)|\le|z|^p,\qquad
|[p](z)|\le\max\{|p||z|,|z|^p\}.
$$
If $0<|z|<\rho_1$, 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 $n\ge2$, put $w=[p](z)$, of order $p^{n-1}$. Since $|z|\ge\rho_1$, the preceding maximum is $|z|^p$. By induction,
$$
|z|^p\ge|w|\ge|p|^{1/(p^{n-1}-p^{n-2})}.
$$
This proves the <formal torsion radius for prime power order>. Taking the positive $p$th root gives
$$
\boxed{|z|\ge|p|^{1/(p^n-p^{n-1})}.}
$$
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