Proof: Let $ψ(k)=p_1p_2...p_k$, where $p_i$ is the $i^{th}$ prime.
Lemma 1 (Local Extrema of Totient Ratio): If $n≤ψ(k)$ then $φ(n)/n ≥ \dfrac{(p_1-1)(p_2-1)...(p_k-1)}{p_1p_2...p_k}$. Proof: Collect $n$ such that $φ(n)/n$ is minimal, and then choose $n=q_1^{α_1}q_2^{α_2}...q_m^{α_m}$ minimal from this collection. Assume $α_i > 1$ for some $i$; then$$φ(n)/n=\dfrac{q_1^{α_1-1}(q_1-1)q_2^{α_2-1}(q_2-1)...q_m^{α_m-1}(q_m-1)}{q_1^{α_1}q_2^{α_2}...q_m^{α_m}}=$$$$\dfrac{(q_1-1)(q_2-1)...(q_m-1)}{q_1q_2...q_m}=φ(n/q_i)/(n/q_i)$$and $n/q_i < n$, violating minimality. So $n$ is squarefree. Let $p_j$ be the smallest prime not dividing $n$, which must be $≤p_k$ else $n=ψ(k)$ and $φ(n)/n$ equals the bound given above. If there is no prime larger than $p_k$ dividing $n$ then set $m=p_jn ≤ ψ(k)$, and otherwise let $q_v$ be this prime and set $m=p_jn/q_v < n ≤ ψ(k)$. In the first case we see $φ(m)/m=\dfrac{p_j-1}{p_j}φ(n)/n < φ(n)/n$ and in the second $φ(m)/m = \dfrac{q_v(p_j-1)}{(q_v-1)p_j}φ(n)/n < φ(n)/n$, invariably violating minimality. Thus the bound holds.$\square$
Lemma 2: $φ(n)→∞$. Proof: Choose finite positive $z$, and let $k$ be such that $(p_1-1)(p_3-1)...(p_k-1) > z$ (index $2$ is missing). We show when $n > ψ(k)$ that $φ(n) > z$. Let $k'$ be such that $ψ(k'-1) < n ≤ ψ(k')$ so that $k' > k ≥ 3$. We observe$$φ(n) = (φ(n)/n)n ≥ \dfrac{(p_1-1)(p_2-1)...(p_{k'}-1)}{p_1p_2...p_{k'}}p_1p_2...p_{k'-1} =$$$$\dfrac{(p_1-1)(p_2-1)...(p_{k'}-1)}{p_{k'}} ≥ (p_1-1)(p_3-1)...(p_{k'-1}-1) ≥$$$$(p_1-1)(p_3-1)...(p_k-1) > z~~\square$$Now, since there are only a finite number of primitive roots for any $n$, $K$ must contain $n^{th}$ primitive roots for $n$ arbitrarily large. Since the degree $φ(n)$ of the cyclotomic minimal polynomial for these primitive roots also becomes arbitrarily large, we must have $K/\mathbb{Q}$ is not finite.$~\square$
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