Let $n = p^em$, and let $N$ be the number of subsets of order $p^e$ in a set of order $n$. Determine the congruence class of $N$ modulo $p$.
I think the answer is m and I feel it is a little similar with the prove of Third Sylow Theorem, which shows the number of Sylow p-subgroups is congruent to 1 modulo p by proving only [H] is fixed by H.
Then, could I decompose the N sets into H-orbit and find m sets that fixed by H in this question?
On
We choose a group $G$ which has order $N$(for example: group of roots of $x^N-1$.)\
S:= set of all subsets of order $p^e$ of $G$.
Take a Sylow-p group of $G$, call $H$.
Introduce $H$ into $S$ by left multiplication.
Then $|S|=\sum \text{orb}(A_i)$ with $\text{orb}(A_i)s $ are all distinct orbits of this operation.
We have: $|\text{orb}(A_i)|=1$ or $p||\text{orb}(A_i)|$ since $\text{orb}(A_i)| |H|=p^e$.
For $A\in S$: $\text{orb}(A)=\{A\} \iff hA=A \forall h\in H$.
Take $a\in A \to Ha \subset A, |Ha|=|A| \to Ha=A \to A \text{ is a right coset of } H. $
Indeed $H$ fixes all right cosets of $H$.
$\to H$ fixes and only fixes right cosets of $H$.
Number of $H$-right cosets =$m$.
Thus, $|S| \equiv m (\mod p)$.
You're trying to compute $\binom{p^{\large e}m}{p^{\large e}}$ mod $p$. Indeed one can prove $\binom{pa+b}{pc+d}\equiv\binom{a}{c}\binom{b}{d}$ mod $p$ using generating functions mod $p$ (binomial theorem!) and then inducting to prove Lucas' theorem, but we can also invoke orbit-stabilizer for a more combinatorially-pure answer.
To prove a set $X$ has a given size mod $p$, one can equip it with a group action of a $p$-group and then show the number of fixed points of the action has that residue.
The cyclic group $C(p^e)$ acts on itself, and thus acts on $C(p^e)\times\{1,\cdots,m\}$ (leaving the second coordinate fixed). Consider the set $\Lambda$ of all subsets of size $p^e$. Show that the fixed points are precisely the sets $C(p^e)\times\{k\}$ for $k\in\{1,\cdots,m\}$, so there are $m$ fixed points.