= Solution
Normalize the <discrete valuations> so a <uniformiser> has value one, and write $k_K,k_L$ for the <residue fields>. The <ramification index> $e$ is specified by $v_L|_K=e\,v_K$, and the <residue degree> is $f=[k_L:k_K]$. For finite extensions of complete discretely valued fields, $[L:K]=ef$. One way to see the degree equality is that $\mathcal O_L$ is a finite free $\mathcal O_K$-module of rank $[L:K]$; reduction modulo a <uniformiser> of $K$ has $e$ successive quotients isomorphic to $k_L$, hence dimension $ef$ over $k_K$.
An <unramified extension> has $e=1$ and separable residue extension, equivalently $[L:K]=f$ with separable residue extension. A <totally ramified extension> has $f=1$, equivalently $[L:K]=e$. The separability condition in the <unramified extension> definition matters if the <residue field> is imperfect; it is automatic for finite <residue fields>.
Suppose $L/K$ is unramified. Choose a <primitive element of a field extension> $\bar x$ for the finite separable extension $k_L/k_K$ and lift it to $x\in\mathcal O_L$. Since the residue degree of $K(x)/K$ is at least $[k_K(\bar x):k_K]=[L:K]$, necessarily $K(x)=L$. Its monic <minimal polynomial of an algebraic element> $f_x$ has coefficients in $\mathcal O_K$. Its reduction has degree $[L:K]$ and annihilates $\bar x$, whose minimal polynomial over $k_K$ has that same degree. The two coincide, so $\bar f_x$ is separable.
Conversely, suppose $L=K(x)$, $x\in\mathcal O_L$, and $\bar f_x$ is separable. It must be irreducible: otherwise its coprime factors lift by the factorization form of <Hensel's lemma>, contradicting irreducibility of $f_x$. Hence $\bar x$ has degree $[L:K]$ over $k_K$. The degree equality forces $f=[L:K]$ and $e=1$, and the residue extension is separable. We have proved the <unramified generator criterion with separable reduction>:
$$
\boxed{L/K\text{ unramified}\iff L=K(x),\ x\in\mathcal O_L,\ \bar f_x\text{ separable}.}
$$
Now let $k_L=\mathbb F_q$, $q=p^f$. Every element of $k_L^\times$ is a simple root of $X^{q-1}-1$, whose derivative is a unit at each root. <Hensel's lemma> lifts each of these $q-1$ elements uniquely to a root in $\mathcal O_L^\times$. Thus \b[$L$ contains all $q-1$ roots of $X^{q-1}-1$]; these are the nonzero <Teichmuller lifts>.
More generally, a <root of unity> of order prime to $p$ reduces injectively into $k_L^\times$. Indeed, if such a root reduces to $1$, uniqueness in <Hensel's lemma> for $X^m-1$ makes it equal to $1$. Consequently every prime-to-$p$ order divides $q-1$.
A root whose order has a nontrivial $p$-part produces a primitive $p$th <root of unity> $\zeta$. It reduces to $1$ in characteristic $p$. Since
$$
p=\prod_{j=1}^{p-1}(1-\zeta^j),\qquad \frac{1-\zeta^j}{1-\zeta}=1+\zeta+\cdots+\zeta^{j-1}\equiv j\pmod{\mathfrak p_L},
$$
all the factors have the same positive integral <valuation>. Hence
$$
e(L/\mathbb Q_p)=v_L(p)=(p-1)v_L(1-\zeta)\geq p-1.
$$
This is the <ramification bound for a primitive pth root of unity>. Therefore
$$
\boxed{e<p-1\implies\mu(L)=\mu_{p^f-1}.}
$$
For $p=2$ the hypothesis $e<1$ cannot occur, so that instance is vacuous rather than a claim excluding the ever-present root $-1$.
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