Since $f(X)=X(\pi+X^{q-1})$, its iterates satisfy
$$
\Phi_m(X):=\frac{f_m(X)}{f_{m-1}(X)}
=\pi+f_{m-1}(X)^{q-1}.
$$
The polynomial $\Phi_m$ is <Eisenstein polynomial>[Eisenstein]: it is monic, every nonleading coefficient is divisible by $\pi$, and its constant term is $\pi$. It is also separable, since $q-1$ is prime to the residue characteristic and the iterates have nonzero derivative.

If $\alpha\ne0$ and $m$ is least with $f_m(\alpha)=0$, then $\Phi_m(\alpha)=0$. Eisenstein irreducibility makes $\Phi_m$ its minimal polynomial, so $K(\alpha)/K$ is totally ramified and separable. For $\alpha=0$ the extension is trivial and has the same properties. This is the <Eisenstein layers of Lubin–Tate torsion> argument.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 136 / 5 / b / Solution

For $f(X)=\pi X+X^q$ and its iterates $f_n$, the quotient
$$
\Phi_n(X)=\frac{f_n(X)}{f_{n-1}(X)}
=\pi+f_{n-1}(X)^{q-1}
$$
is Eisenstein. A nonzero root of exact level $n$ therefore generates a totally ramified separable extension.


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 Incoming links: Eisenstein layers of Lubin–Tate torsion