= Solution
The denominator is $\pi^nX$; dividing by $X$ removes the zero at $X=0$. A <uniformizer> of a degree-$n$ totally ramified extension generates the field: its <valuation> over $K$ is $1/n$, so $K(\pi)/K$ already has ramification index at least $n$. Hence $K(\pi)=L$, its <minimal polynomial> has degree $n$, and its conjugates are the distinct $\sigma(\pi)$ for $\sigma\in G$.
Factor the <minimal polynomial> over $L$. The <reduced ramification polynomial> is
$$
g(X)=\frac{f(\pi(1+X))}{\pi^nX}=\prod_{\sigma\ne1}\left(X-\frac{\sigma(\pi)-\pi}{\pi}\right).
$$
Every displayed root $r_\sigma=(\sigma(\pi)-\pi)/\pi$ is nonzero and integral, since two <uniformizers> have a difference of <valuation> at least one. Thus $g$ is monic of degree $n-1$, has nonzero constant coefficient, and belongs to $\mathcal O_L[X]$. For $n=1$ it is the constant <polynomial> $1$, with no slopes and no ramification jumps.
Here is also a justification of the <uniformizer criterion for lower ramification groups> in this setting. The powers $1,\pi,\ldots,\pi^{n-1}$ are a $K$-basis. In an expansion $x=\sum a_j\pi^j$, the nonzero terms have <valuations> $nv_K(a_j)+j$, distinct modulo $n$; therefore there is no cancellation at the least <valuation>. If $x$ is integral, all those values are nonnegative, which forces $a_j\in\mathcal O_K$. Thus $\mathcal O_L=\mathcal O_K[\pi]$. For $j\geq1$, $\sigma(\pi)^j-\pi^j$ is divisible by $\sigma(\pi)-\pi$, with the remaining factor integral. Consequently every $\sigma(x)-x$ has <valuation> at least $v_L(\sigma(\pi)-\pi)$, and equality is attained at $x=\pi$.
For a nonidentity automorphism, write $i(\sigma)=v_L(\sigma(\pi)-\pi)$. By the defining inequality for the <lower ramification numbering>,
$$
\sigma\in G_m\setminus G_{m+1}\quad\Longleftrightarrow\quad i(\sigma)=m+1\quad\Longleftrightarrow\quad v_L(r_\sigma)=m.
$$
The <Newton polygon root valuation theorem> therefore proves the <ramification breaks from a reduced ramification polygon> relation:
$$
\boxed{-m\text{ is a slope of the coefficient-exponent }N_L(g)\quad\Longleftrightarrow\quad G_m\ne G_{m+1}.}
$$
The segment's horizontal length is exactly $|G_m|-|G_{m+1}|$. Equivalently, in the <reflected Newton polygon convention> required for a positive-slope formulation,
$$
\boxed{m\text{ is a slope of the reflected }N_L(g)\quad\Longleftrightarrow\quad G_m\ne G_{m+1}.}
$$
All occurring root <valuations> are nonnegative integers. For negative indices, extend $G_m=G$; total ramification means that there are no negative-index jumps, so the reflected formulation remains true for all integers.
The sign distinction is substantive. Using the <uniformizer> from question 2, whose <minimal polynomial> is $X^6+3$, gives
$$
g(X)=\frac{(1+X)^6-1}{X}=6+15X+20X^2+15X^3+6X^4+X^5.
$$
Its coefficient <valuations> are $(6,6,0,6,6,0)$, because $v_L(3)=6$. The usual lower hull has vertices $(0,6),(2,0),(5,0)$ and slopes $-3,0$, of lengths $2,3$. These correspond exactly to the drops $|G_3|-|G_4|=2$ and $|G_0|-|G_1|=3$. With the reflected axis the slopes are $0,3$, as in the printed positive-$m$ wording. If the coefficient-exponent definition is used throughout, the printed assertion needs the minus sign shown above.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2016/iii/paper-123-newton-conventions.png]
{title=Coefficient-exponent and reflected Newton polygons of the reduced ramification polynomial}
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