= Solution
Let $\mathcal S$ be the set of all subsets of $G$ with $p^n$ elements. First show that $|\mathcal S|=\binom{p^nr}{p^n}$ is not divisible by $p$. In the <polynomial> ring over the <field> of $p$ elements, $(1+X)^p=1+X^p$, since the intermediate binomial coefficients are divisible by $p$. Iterating gives
$$
(1+X)^{p^nr}=(1+X^{p^n})^r\pmod p.
$$
Comparison of coefficients of $X^{p^n}$ yields $\binom{p^nr}{p^n}\equiv r\not\equiv0\pmod p$.
Let $G$ act on $\mathcal S$ by left translation. Because its <orbit> sizes sum to a number not divisible by $p$, some <orbit> has size prime to $p$. Choose a subset $S$ in that <orbit>, with <stabilizer> $H$. The <orbit-stabilizer theorem> gives the <orbit> size $[G:H]$, so $p\nmid[G:H]$. On the other hand $H$ acts freely on the elements of $S$: if $hs=s$, cancellation in the <group> gives $h=1$. Every $H$-orbit in $S$ therefore has $|H|$ elements, and $|H|$ divides $|S|=p^n$. Since
$$
p^nr=|G|=|H|[G:H]
$$
and the second factor has no factor of $p$, the first must contain all $p^n$. Hence $\boxed{|H|=p^n}$, proving the first of the <Sylow theorems>. This is <Sylow existence by subset action> and includes the trivial case $n=0$.
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