Sylow’s Theorem, from group theory, states that:

1. If $p^k|| |G|$ then $G$ contains a subgroup of order $p^k$.
2. Any two such subgroups are conjugate.
3. The number $n_p$ of such subgroups satisfies $n_p \equiv 1 \mod p$.

The standard proof is to attack 1 using a clever binomial coefficient trick, and then use different arguments to tackle 2 and 3. The following proof, shown to the author by Prof. Leader of Trinity College, Cambridge, more or less proves all three parts simultaneously, and is very natural. The proof is probably slightly different from that I was originally shown, but is hopefully identical in spirit.

Proof: For a finite group $G, p||G|$, let $P$ be a maximal $p$-subgroup (a subgroup of largest cardinality where every element has order a power of $p$).

Proof of part 1: It will suffice if we can prove that $\frac{|G|}{|P|} \not\equiv 0 \mod p$. But $\frac{|G|}{|P|} = \frac{|G|}{|N|} \frac{|N|}{|P|}$, where $N$ is the normaliser of $P$ in $G$. Splitting the fraction in this way allows us to assign meaning to the quantities.

Since $P$ is normal in $N$, we can measure the second quantity as simply the order of the group $N/P$. Suppose $p$ divides this order. Then, by Cauchy’s Theorem, we can find $\bar{H} \leq N/P$ of order $p$. But then $\bar{H}P$ would be a $p$-subgroup and have order greater than $P$, contradicting its maximality.

Now, by the Orbit-Stabiliser theorem considering the action of conjugation on $P$ by $G$, we can see that the first quantity must be the size of the orbit of $P$ under this action. Let us call the set of these orbits $X$, so we need to show that $|X| \not\equiv 0 \mod p$. In fact, we can do better and show that $|X| \equiv 1 \mod p$, which will immediately give part 3 of the theorem as a consequence of part 2.

To do this we consider $P$ acting on $X$ by conjugation, and aim to show that $\{P\}$ is the only singleton orbit (and since $P$ is a $p$-subgroup, all the other orbits must have sizes divisible by $p$, so this would give the required congruence). Suppose $\{gPg^{-1}\}$ is another singleton orbit (with $g \not\in N$). Then $P$ normalises $gPg^{-1}$, which implies $g^{-1}Pg$ normalises $P$, so $g^{-1}Pg \leq N$. But in this case, it is clear that $g^{-1}PgP$ is a $p$-subgroup of order greater than that of $P$ (having nontrivial image of order $p^k$ under the quotient map $N \rightarrow N/P$). Thus the only singleton orbit is $\{P\}$, whence $|X| \equiv 1 \mod p$.

So it remains to prove statement 2 (essentially that $\text{Syl}_p(G) = X$). Suppose $Q$ is a Sylow $p$-subgroup. If we let $Q$ act by conjugation on $X$, then there must be some (unique) singleton orbit $P' = gPg^{-1}$ which is normalised by $Q$. But considering that the image of $QP'$ under the quotient map $N \rightarrow N/P'$ must be a single element (by maximality of $P'$ and the fact all elements of $Q$ have order divisible by $p$), we must have $Q=P'$. So $Q \in X$, as required.

As remarked above, the third part is now immediate.