Induction over membership in the subgroup generated by \(-1\) and the \(\varepsilon _a\): each generator trivially has such an expression, the identity is the empty product, and the set of elements admitting an expression is closed under multiplication (add the exponent vectors) and inverses (negate them).

Let \(x \in C^+\) be a \(p\)th power in \(E^+\), and write \(x = (-1)^s \prod _a \varepsilon _a^{e_a}\) by theorem 2.5. By hypothesis \(p \mid e_a\) for every \(a\). Since \(p\) is odd, \((-1)^s = \bigl((-1)^s\bigr)^p\), so

exhibits \(x\) as the \(p\)th power of an element of \(C^+\).

The group-theoretic statement is: a finite-index \(p\)-saturated subgroup \(H \le E\) whose inclusion captures the \(p\)-torsion of \(E/H\) has index prime to \(p\). Indeed, if \(p\) divided the index, the finite abelian quotient \(E/H\) would contain an element of order exactly \(p\), say the class of \(x\); then \(x^p \in H\) is a \(p\)th power in \(E\), so by saturation \(x^p = y^p\) with \(y\in H\), and \(x/y\) is a \(p\)-torsion unit of \(E\) not in \(H\). For \(E = E^+\), torsion units of the totally real field \(K^+\) are \(\pm 1\), and \(-1 \in C^+\) by construction, so no such element exists — a contradiction. Finiteness of the index is supplied by the cyclotomic-unit index theory (the units \(\varepsilon _a\) are a full-rank family).