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We first prove the simple-root Hensel lemma directly. Put $x_1=x$. Suppose $x_n\equiv x(\mathrm{mod}\pi )$ and $f(x_n)\equiv 0(\mathrm{mod}{\pi }^n)$. The derivative remains a unit since its reduction agrees with $f^{\prime }(x)$. Choose $t_n\in {\mathcal{O}}_K$ whose residue solvesand set $x_{n+1}=x_n+{\pi }^n{t}_n$. Polynomial expansion gives $f(x_{n+1})=f(x_n)+{\pi }^n{t}_n{f}^{\prime }(x_n)+{\pi }^{2n}B$ with $B\in {\mathcal{O}}_K$. Since $2n\ge n+1$, the new value vanishes modulo ${\pi }^{n+1}$. The sequence is Cauchy because $v(x_{n+1}-x_n)\ge n$, and completeness supplies $y\in {\mathcal{O}}_K$ with $y\equiv x(\mathrm{mod}\pi )$. Continuity gives $f(y)=0$.

For uniqueness, if $y,z$ are two roots in that residue class, thenThe factor in parentheses is a unit, since $f^{\prime }(z)$ is a unit and $y-z\in \pi {\mathcal{O}}_K$. Thus $y=z$. This proves existence and uniqueness of the specified lifted root.

Now suppose the residue field has $q=p^r$ elements and put $e=v(p)$. Every element of $k_K^{\times }$ is a simple root of $T^{q-1}-1$. The Hensel lemma therefore lifts its $q-1$ roots uniquely to a subgroup of the roots of unity in $K$. These are the Teichmuller representatives and give all the prime-to-$p$ torsion. To verify the latter assertion, if a root of unity $u\equiv 1$ has order $m$ prime to $p$, write $u=1+t$. In $u^m-1=mt+\left(\genfrac{}{}{0ex}{}{m}{2}\right)t^2+\cdots$, the first term has valuation $v(t)$ and all later terms have greater valuation, so the sum cannot vanish unless $t=0$. Reduction is therefore injective on prime-to-$p$ torsion, whose order divides $q-1$.

Any root of unity has a unique decomposition into a $p$-power-order factor and a prime-to-$p$ factor, by splitting its finite order using coprime powers. For a primitive $p^a$th root $\zeta$, $a\ge 1$, reduction in characteristic $p$ gives $\bar{\zeta }=1$. The cyclotomic identityshows that all factors have the same valuation: $({\zeta }^j-1)/(\zeta -1)=1+\zeta +\cdots +{\zeta }^{j-1}$ reduces to $j\ne 0$. Consequently the valuation of a primitive p-power root of unity isThe left side is a positive integer. Hence $p^{a-1}(p-1)\le e$, so the possible $p$-power orders are bounded. Combining this bound with the prime-to-$p$ result bounds all root orders by one integer, proving finiteness. A finite multiplicative subgroup of a field is cyclic, and it contains exactly the full prime-to-$p$ subgroup of order $q-1$. ThusIf $e<p-1$, even a primitive $p$th root would violate the valuation formula, so $s=0$.

At the boundary $e=p-1$, assume first that $p$ is odd. Take $K_{+}={\mathbb{Q}}_p({\zeta }_p)$. The polynomial ${\Phi }_p(1+T)=p+\left(\genfrac{}{}{0ex}{}{p}{2}\right)T+\cdots +T^{p-1}$ is Eisenstein, so $K_{+}/{\mathbb{Q}}_p$ is totally ramified of degree $p-1$ and $v_{K_{+}}(p)=p-1$. It contains a primitive $p$th root, while the valuation formula excludes primitive $p^2$th roots. Thus $s=1$.

For the other example, take $K_0={\mathbb{Q}}_p(\pi )$ with ${\pi }^{p-1}=p$. The polynomial $T^{p-1}-p$ is Eisenstein, giving the same ramification index and residue field ${\mathbb{F}}_p$. If a primitive $p$th root belonged to $K_0$, the valuation formula would give ${\zeta }_p-1=u\pi$ with $u$ a unit. Substitute this into ${\Phi }_p(1+({\zeta }_p-1))=0$ and divide by $p$. All intermediate terms reduce to zero, leaving $1+{\bar{u}}^{p-1}=0$ in ${\mathbb{F}}_p$. But ${\bar{u}}^{p-1}=1$, contradicting $2\ne 0$. Thus $s=0$ in $K_0$. This proves both cases of roots of unity at the tame ramification boundary for odd primes.

The printed boundary claim needs the qualification $p>2$. When $p=2$, every characteristic-zero field already contains the root $-1$ of order two, so $s=0$ is impossible. If $v(2)=1$, the valuation formula excludes order four and higher, forcing $s=1$. For example $K={\mathbb{Q}}_2$ has exactly this behavior. This is a genuine exception to the final request as printed, rather than an omitted construction.