 Solution

Normalize $v_K$ to have value group $\mathbb{Z}$, and extend it to an algebraic closure. Suppose $\alpha$ is a root of a monic Eisenstein polynomial of degree $n$. A root cannot have nonpositive valuation: the leading term would then have uniquely smallest valuation. If $w=v_K(\alpha )>0$, the constant term has valuation one, all the other nonleading terms have valuation strictly greater than one, and the leading term has valuation $nw$. Cancellation forces $nw=1$. The value group of $K(\alpha )$ is $(1/e)\mathbb{Z}$, so $n\mid e$. Since $e\le [K(\alpha ):K]\le n$, equality holds throughout. Thus the polynomial is irreducible and the extension is totally ramified.

Conversely, let $L/K$ be totally ramified of degree $n$, and choose a uniformizer ${\pi }_L$. Its extended $v_K$ value is $1/n$. The same denominator argument gives $[K({\pi }_L):K]\ge n$, so $K({\pi }_L)=L$. Every conjugate has valuation $1/n$, by the unique extension of an absolute value to a finite extension of a complete field. The nonleading coefficients of its minimal polynomial are elementary symmetric expressions in these conjugates, so their valuations are positive, hence at least one because the coefficients lie in $K$. The constant coefficient is their product and has valuation exactly one. Its minimal polynomial is an Eisenstein polynomial, and $\boxed{L=K({\pi }_L)}$.