The degree of $e^{2\pi i/n}$ goes to infinity with $n$. If $K$ had an infinity of roots of unity, it would have elements of arbitrarily high degree, and thus would not be of finite degree over the rationals, and thus would not, in fact, be an algebraic number field.
I’m not sure what the problem is. You have $K\subset L$, and between them you have the maximal unramified extension of $K$ in $L$, call it $K^{\text u}$. You know that the residue field extensions $f^L_K=f^{K^{\text u}}_K=[\kappa_L:\kappa_K]$, while for the ramification degrees, $e^L_K=e^L_{K^{\text u}}$, i.e. the top layer is totally ramified, the bottom is totally unramified. You have to use multiplicativity of both the residue field degrees $f^?_?$ and the ramification indices $e^?_?$.
If this isn’t enough for you, get back to me in a comment, and I’ll try to fill in any missing details.
EDIT — Addendum:
You have asked why, when adjoining a primitive $q^f$-th root of unity to $K$ you get the maximal unramified extension of $K$ in $L$. Let’s see how this works:
I’ll change your notation slightly, letting the residue-class fields be $\kappa$ and $\lambda$, so that my $\kappa$ is your $\kappa_K$, and my $\lambda$ is your $\kappa_L$. Now, by definition, the residue-field extension degree of $L$ over $K$, which we’re denoting $f$, is $[\lambda:\kappa]$. Let $\zeta_0$ be a primitive ($q^f-1$)-th root of unity
in $\lambda$, say with minimal $\kappa$-polynomial $\varphi(X)\in\kappa[X]$. I’m going to call $q^f=Q$, for convenience.
Now lift $\varphi$ to any monic polynomial $\Phi(X)\in K[X]$; this is still $K$-irreducible. Since $\varphi$ splits into linear factors over $\lambda$, one of these being $X-\zeta_0$,
Hensel tells us that $\Phi(X)$ also splits into linear factors over $L$, one of these being $X-\xi$ with $\xi\mapsto\zeta_0\in\lambda$, but of course $\xi$ need not be a root of unity at all.
Nonetheless, $\xi^{q^f}\equiv\xi\mod{\mathfrak M}$, where $\mathfrak M$ is the maximal ideal in the integers of $L$. I’ll ask you to consider the sequence
$$\xi, \xi^Q, (\xi^Q)^Q=\xi^{Q^2},\xi^{Q^3},\cdots$$
which you easily show is $\mathfrak M$-adically convergent, all of the terms being congruent to $\xi$ modulo $\mathfrak M$. And of course the limit is a ($Q-1$)-th root of unity reducing to $\zeta_0$ in $\lambda$.
This limit is what I’ll call $\zeta$; it’s a primitive such root. Its
minimal polynomial is congruent to $\Phi$ modulo $\mathfrak M$, that is, it reduces to $\varphi(X)$, and it’s still irreducible.
Putting it all together, what do we have? $K(\zeta)=K(\xi)$ is of degree $f$ over $K$, unramified since its residue-field degree is equal to its vector-space degree. And certainly there is no larger unramified extension
of $K$ in $L$. So I think that you have your maximal unramified extension, generated by a root of unity of the right type.
Best Answer
As you noticed, the first term ${n \choose 1}x$ dominates, unless $p\mid n$. Actually any root of unity $u=1+x$ in $U^{(1)}$ must be of order that is a power of $p$, for otherwise a suitable power of $u$ would have order prime to $p$, and we just saw that this cannot happen.
Such roots of unity may or may not exist in your field. For example the $p$-adic field $\mathbf{Q}_p$ has no roots of unity of order $p$, if $p>2$, because the term ${p\choose 1}x$ dominates (all the binomial coefficients save the last one are divisible by $\pi=p$ exactly once, and also $p\mid x$, so that last term $x^p$ is then also divisible by a higher power of $p$ than the first term).
But surely you can adjoin a $p$th root of unity $\zeta_p$ to $\mathbf{Q}_p$! If you do that you get a ramified extension of $\mathbf{Q}_p$. You have probably seen the same thing happen with the cyclotomic extension $\mathbf{Q}(\zeta_p)/\mathbf{Q}$ of algebraic number fields. There the prime $p$ is totally ramified.
Note also that the above argument showing that $(1+x)^p\neq1$ for a non-zero $x\in p\mathbf{Z}_p$ fails, when there is ramification. The last term $x^p$ may then be divisible by the same power of $\pi$ as the ( in the earlier case dominating) first term $px$. In the case $\mathbf{Q}_p(\zeta_p)$ we can thus deduce that $p$ must be divisible by $\pi^{p-1}$.