= Solution
Put $C=\bigcup_{r\geq1}\mathbb Q(\mu_{p^r})$, where $\mu_{p^r}$ is the group of $p^r$th <roots of unity>. The compatible <Galois groups> of the <cyclotomic fields> give
$$
\operatorname{Gal}(C/\mathbb Q)
\cong\varprojlim_r(\mathbb Z/p^r\mathbb Z)^\times
=\mathbb Z_p^\times.
$$
For odd $p$, the <Teichmüller character> gives
$$
\mathbb Z_p^\times=\mu_{p-1}\times(1+p\mathbb Z_p).
$$
The <p-adic logarithm> identifies $1+p\mathbb Z_p$ with $p\mathbb Z_p$, which is topologically isomorphic to $\mathbb Z_p$. More explicitly, $a\mapsto(1+p)^a$ is an <isomorphism> from the additive <p-adic integers> onto $1+p\mathbb Z_p$: the logarithm of $1+p$ has <valuation> one and generates $p\mathbb Z_p$. Take the fixed field of $\mu_{p-1}$ in $C$.
For $p=2$, use instead
$$
\mathbb Z_2^\times=\{\pm1\}\times(1+4\mathbb Z_2).
$$
The <p-adic logarithm> identifies the second factor with $4\mathbb Z_2$, and $a\mapsto5^a$ identifies it with $\mathbb Z_2$. The fixed field of $\{\pm1\}$ in $C$ is the union of the maximal real subfields of the <cyclotomic fields> of $2$-power conductor. Thus in both cases the \b[cyclotomic Zp-extension] exists:
$$
\boxed{\mathbb Q[p^\infty]=C^{\Delta_p},\qquad
\operatorname{Gal}(\mathbb Q[p^\infty]/\mathbb Q)\cong\mathbb Z_p,}
$$
where $\Delta_p=\mu_{p-1}$ for odd $p$ and $\Delta_2=\{\pm1\}$.
The closed subgroup $p^n\mathbb Z_p$ is the unique subgroup of index $p^n$, so the <Galois correspondence> supplies the unique degree-$p^n$ field $\mathbb Q[p^n]$. In the odd case it is the $\Delta_p$-fixed field in $\mathbb Q(\mu_{p^{n+1}})$; in the even case it is the real subfield of $\mathbb Q(\mu_{2^{n+2}})$. These finite layers are all <totally real number fields>.
Now set $A=\mathbb Q[p^n]$, $B=\mathbb Q[p^m]$, and let $H_A$ be the ordinary <Hilbert class field> of $A$. Thus $H_A/A$ is abelian, unramified at finite primes, and split at every real place, with degree $h(A)$. If $m=n$ there is nothing to prove. Otherwise the unique prime over $p$ is totally ramified in $B/A$, by the permitted total-ramification fact and multiplicativity of the <ramification index>. Any nontrivial intermediate field of $B/A$ is ramified at that prime. Hence
$$
H_A\cap B=A.
$$
It follows that the two extensions are <linearly disjoint field extensions>, and
$$
[BH_A:B]=[H_A:A]=h(A).
$$
Unramified extensions remain unramified after base change. The real places also remain split: $H_A$ and $B$ are totally real, so their compositum is totally real. Therefore $BH_A/B$ lies inside the ordinary <Hilbert class field> $H_B/B$. Its degree divides $[H_B:B]=h(B)$, proving
$$
\boxed{h(p^n)\mid h(p^m).}
$$
This is <class number divisibility with split real places>; using a class-field statement restricted to imaginary fields would unnecessarily omit the present real layers.
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