We use group actions and counting modulo $p$. Let $\mathcal{S}$ be the set of all subsets of $G$ of size $p^n$: $\mathcal{S} = \{S \subseteq G : |S| = p^n\}$

The size of $\mathcal{S}$ is the binomial coefficient: $|\mathcal{S}| = \binom{p^n m}{p^n}$

Step 1: Show $|\mathcal{S}| \not\equiv 0 \pmod{p}$

Write $\binom{p^n m}{p^n} = \frac{(p^n m)!}{(p^n)!(p^n m - p^n)!}$. By Legendre's formula, the power of $p$ dividing this binomial coefficient is: $v_p\left(\binom{p^n m}{p^n}\right) = v_p((p^n m)!) - v_p((p^n)!) - v_p((p^n(m-1))!)$

Careful calculation shows this equals 0, so $|\mathcal{S}| \not\equiv 0 \pmod{p}$.

Step 2: Define a group action

Let $G$ act on $\mathcal{S}$ by left multiplication: $g \cdot S = \{gs : s \in S\}$

This partitions $\mathcal{S}$ into orbits. By the orbit-stabilizer theorem, each orbit size divides $|G|$.

Step 3: Find an orbit not divisible by $p$

Since $|\mathcal{S}| \not\equiv 0 \pmod{p}$ and $|\mathcal{S}|$ is the sum of orbit sizes, at least one orbit $\mathcal{O}$ must satisfy $|\mathcal{O}| \not\equiv 0 \pmod{p}$.

Step 4: Show the stabilizer has order $p^n$

For any $S_0 \in \mathcal{O}$, we have: $|\mathcal{O}| = [G : \text{Stab}(S_0)] = \frac{p^n m}{|\text{Stab}(S_0)|}$

Since $|\mathcal{O}| \not\equiv 0 \pmod{p}$ and $p^n m$ has exactly $n$ factors of $p$, we must have $|\text{Stab}(S_0)| = p^n k$ for some $k$ dividing $m$. But $|\mathcal{O}|$ must divide $p^n m$, forcing $|\text{Stab}(S_0)| \geq p^n$.

By a counting argument involving how $\text{Stab}(S_0)$ acts on $S_0$, one shows $|\text{Stab}(S_0)| \leq p^n$. Thus $|\text{Stab}(S_0)| = p^n$.