In my algebra textbook, it goes like this. First, there is presented Cauchy's theorem:
Let $G$ be a finite group, and let $p$ be a prime divisor of $|G|$. Then $G$ contains an element of order $p$.
Then goes the proof:
Consider the set
$S = \{(a_1, ..., a_p)| a_1...a_p = e_G \} \subseteq G^p$
Once $a_1, ... a_p$ are chosen (arbitrarily), then $a_p$ is determined is determinaed uniquely as the inverse of $a_1...a_{p-1}$. Then $|S| = |G|^{p-1}$, so $p$ divides the order of $S$ as it divides the order of $G$.
Now we may act with the group $\mathbb{Z}/p\mathbb{Z}$ on $S$: given $[m]_p \in \mathbb{Z}/p\mathbb{Z}$, $0 \leq m < p$, act by $[m]_p$ on
$(a_1, ..., a_p)$
by sending it to
$(a_{m+1}, ..., a_p, a_1, ..., a_m)$, which is still an elements of $S$.
Now we know that $|Z| \equiv |S| \equiv 0 \mod p$,
where $Z$ is the set of pixed points of this action. That is, $Z = \{ (a, ..., a)| a^p = e_G \} \subseteq G^p$. Note that $Z \neq \varnothing$,since $(e_G, ..., e_G) \in Z$. Since $p \geq 2$ and $p$ divides $|Z|$, we conclude that $|Z| > 1$; therefore there exists some element in $Z$ of the form $(a, ..., a)$ with $a \neq e_G$. Then $\exists \ \ a \in G \setminus \{e_G \}: \ \ a_p = e_G$, then the order of $a$ is $p$.
Now the reader is asked to prove the following claim:
Let $G$ be a finite group, let $p$ be a prime divisor of $|G|$, and let $N$ be the number of cyclic subgroups of $G$ of order $p$. Then $N \equiv 1 \mod p$.
I have a problem now. Isn't $N = |Z| - 1$? Because we don't count $e_G$. If $|Z| = pk, k \in \mathbb{N}$, then $N = |Z| - 1 = pk - 1$, so $N \equiv -1 \mod p$.
Could you, please, point at my mistake? Can't see it at the moment, even though I suspect it might be obvious.
Your equation $\lvert N \rvert=\lvert Z\rvert-1$ is wrong. Not every element of $Z$ maps to a distinct cyclic subgroup.
For every cyclic subgroup of order $p$, we have $e_G$ and then we have $p-1$ other elements of order $p$. Thus, for each cyclic subgroup, we add $p-1$ elements to $Z$. Therefore, $Z$ has $1$ element for $e_G$ and $p-1$ elements for each cyclic subgroup. If we say the number of cyclic subgroups are $N$, then we get: $$\lvert Z \rvert=1+(p-1)\lvert N \rvert$$ As you said, we should substitute $pk$ for $\lvert Z\rvert$: $$pk=1+(p-1)\lvert N \rvert$$ Solve for $\lvert N \rvert$: $$\lvert N\rvert=\frac{pk-1}{p-1}$$ Both the numerator and the denominator are $\equiv -1 \pmod p$, so we find that $\lvert N\rvert \equiv (-1)(-1)^{-1} \equiv 1 \pmod p$.