= Solution
A <place of a number field> is an equivalence class of nontrivial <absolute values on a field>. Its finite places correspond to nonzero prime ideals of the <ring of integers of a number field>; its infinite places come from real embeddings and conjugate pairs of complex embeddings.
For a <embedding of a number field into a p-adic algebraic closure> $\iota:K\hookrightarrow\overline{\mathbb Q}_p$, pull back the <p-adic absolute value>. This gives a place above $p$. Every element of $\operatorname{Gal}(\overline{\mathbb Q}_p/\mathbb Q_p)$ preserves that absolute value, by uniqueness of its extension to each finite local extension, so equivalent embeddings give the same place.
Conversely, a place $v$ above $p$ gives a <completion of a number field at a prime ideal> $K_v$, a finite extension of $\mathbb Q_p$. Embed it into $\overline{\mathbb Q}_p$ and restrict to $K$. If two embeddings give the same place, they extend to two $\mathbb Q_p$-embeddings of this same completion. They are conjugate under the absolute <Galois group>: their finite separable images lie in a finite normal closure, and an isomorphism of such images extends to an automorphism of the algebraic closure. The completion really is the closure of the embedded $K$, since rational coefficients are dense in the $\mathbb Q_p$-span of a primitive element. Thus we obtain the <embedding orbit description of finite places>:
$$
\boxed{\{v\mid p\}\ \longleftrightarrow\ \operatorname{Gal}(\overline{\mathbb Q}_p/\mathbb Q_p)\backslash\operatorname{Emb}(K,\overline{\mathbb Q}_p).}
$$
For the <product formula>, use normalized local factors
$$
|x|_{\mathfrak p}=(N\mathfrak p)^{-\operatorname{ord}_{\mathfrak p}(x)},\qquad
|x|_v=|\sigma(x)|\text{ at a real place},\qquad
|x|_v=|\sigma(x)|^2\text{ at a complex place}.
$$
The squared modulus at a complex place counts its two embeddings. Equivalently one can use ordinary complex modulus and put exponent two in the product. These <normalized local factors for the product formula> must be specified; arbitrary representatives of place classes would not satisfy the unweighted printed formula.
The fractional <principal ideal> $(x)$ has only finitely many nonzero prime exponents. Its <ideal norm> is $|N_{K/\mathbb Q}(x)|$: for integral $x$, multiplication by $x$ on an <integral basis> has this determinant in absolute value, equal to the index of $(x)$ in $\mathcal O_K$; a quotient gives the fractional case. Hence the finite-place product is $|N_{K/\mathbb Q}(x)|^{-1}$. The infinite-place product is $|N_{K/\mathbb Q}(x)|$, by the embedding expression for the <field norm>. Therefore \b[all but finitely many local factors are one], and
$$
\boxed{\prod_v|x|_v=1.}
$$
Finally, let $\Delta_K\ne0$ be the integer <field discriminant>, defined by the determinant of the <trace pairing> on an <integral basis>. If a rational prime $p$ ramifies, the algebra $\mathcal O_K/p\mathcal O_K$ has a nonzero nilpotent element: use <prime ideal factorization> and the <Chinese remainder theorem> to see a nonzero nilpotent in a factor $\mathcal O_K/\mathfrak p^e$ with $e>1$. If $y$ is nilpotent, multiplication by $yz$ is nilpotent for every $z$ in this commutative algebra, and therefore has trace zero. Thus $y$ is in the radical of its <trace pairing>, making the reduced discriminant zero. We have proved the <discriminant obstruction to ramification>:
$$
p\text{ ramifies}\implies p\mid\Delta_K.
$$
Only finitely many rational primes divide this nonzero integer, so \b[only finitely many primes ramify].
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