Let $P \in Syl_{p}(G)$. Show that for every nonnegative integer $a$, the number of subgroups of size $p^a$ in $P$ is congruent mod $p$ to the number of size $p^a$ subgroups in $G$.
The text says that this leads to an alternative proof that $n_p(G)\equiv1 \pmod p$, so I am trying not to use that fact. I am pretty stumped on this problem, though I was able to think about what each congruence is. I am missing whatever steps show that they're equal.
For example, if $Q$ is a subgroup of $P$ that acts on the set of subgroups of size $p^a$ in $P$ by conjugation, then the size of the orbit of any subgroup must be a power of $p$, unless $Q$ fixes it, in which case the orbit has size 1. This shows that the number of size $p^a$ subgroups in $P$ is congruent mod $p$ to the number of subgroups of size $p^a$ in $P$ that are fixed by conjugation in $Q$.
The same is true if we replace $P$ in the last paragraph with $G$. Here $Q$ is potentially fixing more subgroups outside of $P$, but if we can show that the number of subgroups that get fixed that are outside of $P$ is divisible by $p$, (no idea if this is doable) then the problem is finished.
At this point I am unsure of how to proceed. I would imagine this is how a solution starts, since a lot of results in the book related to Sylow theory are shown by looking at the conjugation action of the right subgroup, so maybe this requires the right choice of $Q$. Some inductive argument would also make sense since this is trivial for $a=0$ and a proof for the case where $p^a=|P|$ is also conceivable. I would really appreciate any nudge in the right direction.
Edit: Also, counting the total number of such subgroups is aided by the fact that Sylow p-subgroups are isomorphic and therefore contain the same number, but I haven't been able to think of a way that this would help since the Sylow p-subgroups can intersect in so many different ways and I don't have a way of counting the number of Sylow p-subgroups.
Found some help here.
If the action described in the question comes from $P\in Syl_p(G)$ and $S$ is some p-subgroup that conjugation with P fixes, then $SP=PS$ and so $SP$ is a subgroup. It is easy to see (and is an earlier exercise in the text) that $S\subset P$. The number of subgroups of size $p^a$ that are fixed by conjugation from $P$ is $0$, which from the reasoning above finishes the problem.