Total ramification $p = \epsilon \pi^n$ implies $F = K(\pi)$ and minimal polynomial of $\pi$ Eisenstein Before I state my question, let me give the set-up / "what I know".
Let $F / K$ be an extension of number fields of degree $n$.  Let $v$ be a (discrete) valuation on $K$ and let 
$$
\mathfrak{O}_{v} = \{ \alpha \in K : v(\alpha) \geq 0  \}
$$ 
be the valuation ring of $v$.  Up to associates,  $\mathfrak{O}_{v}$ has a unique irreducible element $p$ and $v(p) = 1$.  
Let $v_1, \ldots, v_m$ be the extensions of $v$ to the field $F$, let 
$$
\mathfrak{O}_{v_i} = \{ \alpha \in F : v_i(\alpha) \geq 0  \}
$$ 
be the valuation ring of $v_i$ $(i = 1, \ldots, m)$, and let $\pi_i$ be the unique (up to associates) irreducible element of $\mathfrak{O}_{v_i}$ $(i = 1, \ldots, m)$.  Then the integral closure of $\mathfrak{O}_{v}$ in $F$ is
$$
\overline{\mathfrak{O}}_{v} = \cap_{i=1}^{m} \mathfrak{O}_{v_i},
$$
the elements $\pi_1, \ldots, p_m$ constitute all the nonassociate irreducibles in $\overline{\mathfrak{O}}_{v}$, and $p$ factors in $\overline{\mathfrak{O}}_{v}$ as
$$
p = \epsilon \pi_{1}^{e_1} \cdots \pi_{m}^{e_m} 
$$
where $\epsilon$ is a unit of $\overline{\mathfrak{O}}_{v}$, and where $e_i$ is the ramification index of $v_i$ with respect to $v$ (i.e., $e_i$ is the positive integer such that $v(\alpha) = v_i(\alpha)/e_i$ for all $\alpha \in K^{\times}$, i.e., $e_i = v_i(p)$).
The residue class field of $v$ is $\Sigma_{v} = \mathfrak{O}_v / m_v$, where 
$$
m_v = (p) = \{ \alpha \in K : v(\alpha) > 0  \}
$$ 
is the unique maximal ideal of $\mathfrak{O}_v$.  Similarly, the residue class field of $v_i$ is $\Sigma_{v_i} = \mathfrak{O}_{v_i} / m_{v_i}$, where 
$$
m_{v_i} = (\pi_i) = \{ \alpha \in F : v_i(\alpha) > 0  \}
$$ 
is the unique maximal ideal of $\mathfrak{O}_{v_i}$.  The residual degree of $v_i$ over $v$ is $f_i =  [ \Sigma_{v_i} : \Sigma_{v} ]$
The following formula relating the ramification indices $e_i$, the residual degrees $f_i$, the number $m$ of extensions of $v$ to $F$, and the degree $n$ of $F / K$ holds:
$$
\sum_{i=1}^{m} e_i f_i = n
$$
Now the question: Suppose that $p$ factors in $\overline{\mathfrak{O}}_{v}$ as
$$
p = \epsilon \pi^n
$$
where $\epsilon$ is unit in $\overline{\mathfrak{O}}_{v}$ and $\pi$ is an irreducible element in $\overline{\mathfrak{O}}_{v}$.  
Part 1: I want to show that $F = K(\pi)$ (that is, $\pi$ is a primitive element for $F/K$).  To do this, it suffices to show the minimal polynomial of  $\pi$ over $K$, 
$$
\phi(X) = X^k + a_{k-1}X^{k-1} + \cdots + a_0 \qquad (a_i \in K),
$$
has degree $n$.  Of course, $k \leq n$.  
Alex Bartel has offered a solution below, but I don't understand some of the details.  Here is my understanding of his argument.  By definition, 
$$
0 = \phi(\pi) = \pi^k + a_{k-1}\pi^{k-1} + \cdots + a_0
$$
On the left, for every $i$, $v_i(\phi(\pi)) = v_i(0) = \infty$.  On the right, $v_i(\pi^k) = k$, $v_i(a_{j}\pi^{j}) = v_i(a_{j}) + j = nv(a_j) + j$ for $j=0, \ldots, k-1$, and so
$$
v_i(\phi(\pi)) \geq \min(v_i(\pi^k), v_i(a_{j}\pi^{j})) = \min(k, nv(a_j) + j)
$$
where $j$ ranges from $0$ to $k-1$.  If $k \neq nv(a_j) + j \neq nv(a_l) + l$ for all $j \neq l$, then equality holds in the last displayed equation, which contradicts that the left-hand side is infinite.  Alex asserts that $k \neq nv(a_j) + j \neq nv(a_l) + l$ for all $j \neq l$ holds if $k < n$, meaning $k < n$ is impossible.  I do not understand why this assertion should be ture  
Part 2: I also want to show that the minimal polynomial of $\pi$ over $K$,
$$
\phi(X) = X^n + a_{n-1}X^{n-1} + \cdots + a_0 \qquad (a_i \in K),
$$
is Eisenstein with respect to $v$ (that is, $v(a_{i}) > 0$ for $i = 1, \ldots, n-1$ and $v(a_0) = 1$ [I think this is what Eisenstein with respect to $v$ means]).
Based on the formula $\sum_{i=1}^{m} e_i f_i = n$, I know that $v$ has only one extension to $F$, $v_1$, with $e_1 = n$, $f_1 = 1$, and irreducible element $\pi_1 = \pi$.  So $\pi$ is the unique irreducible element of $\overline{\mathfrak{O}}_{v}$ and every nonzero element $\alpha$ in $\overline{\mathfrak{O}}_{v}$ factors as
$$
\alpha = \epsilon \pi^{a}
$$
for some unit $\epsilon$ in $\overline{\mathfrak{O}}_{v}$ and some $a \geq 0$.  Since the quotient field of $\overline{\mathfrak{O}}_{v}$ is $F$, any nonzero $\alpha$ in $F$ can be expressed as above if we allow  $a$ to range through the integers.
That is where I am stuck.
I would like to solve this problem in the language of valuations as I am studying from Borevich and Shafarevich, and this is the language they use.
I would appreciate some help with this.
 A: First, to show that $F=K(\pi)$, it is enough to show that $K(\pi)$ has degree $n$ over $K$, since one inclusion is clear. In fact, you only need to show that $K(\pi)$ has degree at least $n$. Let $v$ be the normalised valuation on $K$ corresponding to $p$ (which by the way means that $v(p)=1$, not 0 as you wrote), and extend $v$ to $F$, so that $v(\pi)=1/n$. Let $f=\sum_{i=1}^m a_mX^m$ be the minimal polynomial of $\pi$ over $K$. If $m<n$, then the valuations of all the monomials in $f(\pi)$ are distinct, since they are distinct modulo $\mathbb{Z}$, so $f(\pi)\neq 0$. You deduce that we must have $m=n$ and this settles your first question.
To show that $f$ is Eisenstein, you again consider valuations of the monomials. As we have already observed, if some monomial in $f(\pi)$ has valuation strictly smaller than all the others, then $f(\pi)\neq 0$. Since $v(\pi^n) = 1$, you deduce that $v(a_i)\geq 1$ for all $i$. But if $v(a_0)>1$, then the top power term has the lowest valuation. which we agreed couldn't be the case, so you are done.
