Quickest and/or most elementary proof of "principal iff splits completely"? 
Let $L$ be the Hilbert class field of a number field $K$, and let $\mathfrak{p}$ be a prime ideal of $K$. Then $\mathfrak{p}$ splits completely in $L$ if and only if $\mathfrak{p}$ is a principal ideal.

What is the quickest and/or most elementary proof of this fact, and if so, where can I find it? I understand that the standard proof requires quite a bit of machinery from class field theory, and I'm hoping to streamline that process...
EDIT: So as the comment by Franz Lemmermeyer suggests, we ask the alternative question:

What is the quickest proof of the equivalence between Weber's and Takagi's definition of a class field?

 A: Let $K/F$ be an extension of number fields. We say that $K/F$ is 


*

*a Weber-Hilbert class field if the prime ideals of $F$ that split 
completely in $K$ are exactly the principal prime ideals.

*a Takagi-Hilbert class field if the norms of all ideals from $K$
are principal ideals in $F$: $N_{K/F} D_K \subseteq P_K$, where
$D_K$ is the group of nonzero fractional ideals in $K$ and $P_K$ 
the group of nonzero fractional ideals in $F$. 

*an Artin-Hilbert class field if $K/F$ is abelian and if the Artin
symbol induces an exact sequence 
$$ 1 \longrightarrow P_K \longrightarrow D_K \longrightarrow 
     Gal(K/F) \longrightarrow 1 $$
Since these definitions involve ideals rather than ideles my guess is
that the most simple proof of the equivalence of these definitions
uses the classical approach. This approach in turn uses analytic tools,
and for these tools sets of prime ideals with density $0$ are invisible.
Thus in the classical theory, which is based on Takagi's definition, 
the equivalence with Weber's definition for all ideals (and not just
all except for a set of density $0$) is proved after all the main
theorems.
What is clear from the definitions is that if $K/F$ is a Takagi-Hilbert
class field, then each prime ideal that splits completely in $K/F$ must 
be principal. The proof that in a Takagi-Hilbert class field exactly the 
princiapal primes split completely up to a set of exceptions with 
density $0$ may be accomplished using analytic tools and should not 
be too difficult. 
It is also clear that an Artin-Hilbert class field is a Weber-Hilbert
class fields, since prime ideals have trivial Artin symbol if and only
if they split completely. For proving the converse you have to show that
a Weber-Hilbert class field is abelian. Arnold Scholz ( Zur 
Klassenkörpertheorie auf Takagischer Grundlage, Math. Z. 30
 (1929), 332--356) has shown that Takagi-Hilbert class fields must be
abelian, and perhaps this proof may be transferred to Weber-Hilbert
class fields. 
You will get a proof that Takagi-Hilbert class fields are Weber-Hilbert 
class fields by taking the classical approach to class field theory
based on Takagi's definition and following the proof up to the 
decomposition law, restricting everything to unramified extensions 
(which simplifies a few things, but not dramatically). If anyone
has a better idea, I'm all ears.
Addition. It is of course also possible to define the Hilbert class field $H$ 
of a number field $K$ as the maximal abelian unramified extension of $K$. 
A naive idea for showing that exactly the principal prime ideals
split completely in $H/K$ would be trying to prove this in its
cyclic subextensions; but this is bound to fail: if $K$ has class
group of type $(2,2)$, then the prime ideals from a class of order $2$
remain inert in two of the three quadratic unramified extensions.
Thus for proving the decomposition law you will have to use the
maximality condition, which brings me to the first question:
 Is there a simple proof that the maximal abelian unramified 
     extension of a number field is finite? 
If one could generalize Hilbert's Satz 94 to noncyclic extensions I 
would guess that the answer to this question is yes. The main ingredients 
of such a proof certainly would be Dirichlet's unit theorem (finite 
generation of the unit group) and the finiteness of the class number. 
Or is there a different way of linking unramified abelian extensions to 
the class group?
The classical proof of the finiteness of the Hilbert class field runs as
follows. Let $L/K$ be a cyclic unramified extension, and consider the index
$$ h_{L/K} = (D_K : ND_L \cdot P_K). $$
Using elementary transformations and the ambiguous class number formula
(essentially the Herbrand unit index calculation) this can be shown to equal
$$  h_{L/K} = (L:K) (E_K : E_K \cap NL^\times) (Cl(L)[N] : Cl(L)^{1-\sigma}), $$
where $\sigma$ generates the Galois group of $L/K$.
By the first inequality (a consequence of the fact that the Dedekind zeta
function has a pole of order $1$ at $s = 1$) we must have $h_{L/K} \le (L:K)$
for all cyclic unramified extensions, from which we conclude that 
$h_{L/K} = (L:K)$, and that 
$$(E_K : E_K \cap NL^\times) = (Cl(L)[N] : Cl(L)^{1-\sigma}) = 1$$ 
for all cyclic unramified extensions $L/K$. From here it is not difficult to 
prove that $h_{L/K} = (L:K)$ holds for all abelian unramified extensions. 
This in turn proves that $(L:K) = h_{L/K} \le h_K$ for any abelian 
unramified extension.
I do not think that this inequality is sufficient for getting hold
of the decomposition law. If the maximal unramified abelian extension 
of $K$ has degree $< h_K$ then you are in trouble. If, for example,
${\mathbb Q}(\sqrt{-5})$ were its own Hilbert class field, all ideals
would split in $H/K$, not just the principal ones. Thus I fear 
that for proving the decomposition law you actually need the
existence of an unramified abelian extension of degree $h_K$.
If this is true, then I guess that there really is no simple proof
of the decomposition law: the existence theorem in class field theory
is a technical tour de force even (and perhaps even more so) if you
restrict your attention to unramified extensions.
