Let $K$ be a totally ramified degree $p^n$ extension of $\mathbb{Q}_p$. Write $G_{K/\mathbb{Q}_p}$ for the set of $\mathbb{Q}_p$-algebra homomorphisms from $K$ into some fixed algebraic closure, so that in the Galois case, $G_{K/\mathbb{Q}_p} = \operatorname{Gal}(K/\mathbb{Q}_p)$. In this paper, Lbekkouri defines the lower ramification number $v_\sigma$ of $\sigma \in G_{K/\mathbb{Q}_p}$ to be $$ v_\sigma = v\Big(\frac{\sigma(\pi) - \pi}{\pi}\Big), $$ where $\pi$ is a uniformiser of $K$. The $v_\sigma$ are precisely the integers $i$ such that $G_i/G_{i+1}$ is nontrivial, where the $G_i$ are the ramification subgroups of $G_{K/\mathbb{Q}_p}$. These quotients are subgroups of $U_i/U_{i+1} \cong \mathbb{Z}/p\mathbb{Z}$, where $U_i = 1 + (\pi^i)$, so exactly $n$ of them are nontrivial, hence the $v_\sigma$ take $n$ distinct values as $\sigma$ ranges over $G_{K/\mathbb{Q}_p}$. Let us order these values as $v_0 < v_1 < \ldots < v_{n-1}$.
Question: Lbekkouri claims that we have $v_r \leq \frac{p^{r+1}}{p-1}$ for each $r$. Why is this true?
My attempt: I have tried to solve the problem for $r=0$ and $p$ odd. That is, I want to show that $v_0 \leq \frac{p}{p-1}$. Since I am assuming $p$ odd, I need $v_0 \leq 1$. In other words, we want some $\sigma$ such that $v(\sigma(\pi) - \pi) \leq 2$, which is equivalent to $$ \sigma(\pi) \not \equiv \pi \pmod{\pi^3} $$ for some $\sigma$. So basically, the roots of the minimal polynomial $f(X)$ of $\pi$ can't be too close together. Somehow, I want to extract a contradiction from the assumption that $\sigma(\pi) \equiv \pi \pmod{\pi^3}$ for all $\sigma$. Assuming that this is the case, we can write $$ f(X) = \prod_{i=1}^{p^n}(X - \pi - \pi^3\theta_i) $$ for integers $\theta_i$ of the algebraic closure of $K$. I have considered the coefficients of $X^{p^n-1}$ and $X^0$ in this polynomial to try and find a contradiction (since we know these must be in $\mathbb{Z}_p$, and that $f(X)$ is Eisenstein), but I haven't been successful. I also though about trying to apply Hensel's Lemma to $$ f(X) \equiv (X - \pi)^{p^n} \pmod{\pi^3}, $$ but again I haven't managed to find a contradiction. My feeling is that somehow, we might be able to leverage the proximity of the roots to "lift the degeneracy" from $f \pmod{\pi^3}$ to $f$.