Proving that $a$ is a $p$-cycle

Question

I was reading Topic in Algebra by I.N. Herstein and trying to solve a problem from it.

If $p$ is a prime number, show that in $S_p$ there are $(p-1)!+1$ elements $x$ satisfying $x^p=e$.

I was able to solve it. Here is how I did it.

By Caychy's theorem as $p\mid o(G)\color{grey}{=p!}$ there exists an element of order $p$ and not equal to $e$. Call it $a$

Let $G=\langle a\rangle$, clearly $G$ is a subgroup of $S_p$. It is easy to see that every element of conjugate class of $a$ satisfies $k^p=e$. Using Cayley's theorem, $G$ is isomorphic to a finite group of permutations. Recalling the proof of Cayley's theorem, the set is given by $\{\tau_g|g\in G\}$. It is easy to see that $\tau_a=(1,2,3,\cdots,p)$.

Now I can use that $|C_a|=|C_{\tau_a}|=(p-1)!$. And $e$ is a trivial solution to $x^p$ and is not included in the conjugate class of $a$.

---

I think that using Cayley's theorem is an overkill, especially when I'm applying it to the group of permutations itself. I also think that we can prove that $a$ is a $p$ cycle which completes the proof immediately.

How can we prove that $a$ is a $p$ cycle?

Answer 1

Hint:

An element in $\;S_n\;$ has order $\;q=$ a prime iff it is the product of disjoint $\;q-$cycles. Now, in $\;S_p\;$ how many disjoint $\;p-$cycles are there?

Answer 2

The elements which satisfy the above equation are only $p$ cycles and product of disjoint $p$ cycles, but there is no permutation which is product of disjoint $p$ cycles and there are $(p-1)!$ $p$-cycles, so there are $1+(p-1)!$ elements satisfying the equation as identity also satisfy the equation.