Proof validation : Number of solutions to $x^m = e$ is $m$, where $x \in G$ and $|G| = n$ I'm currently going through Fraleigh's exercises in "A First Course in Abstract Algebra, 7 Ed", in particular Ex 6.53:

Show that in a finite cyclic group $G$ of order $n$, written manipulatively, the equation $x^m = e$ has exactly $m$ solutions $x$ in $G$ for each positive integer $m$ that divides $n$.

My solution is as follows:

Let $G = \left<a\right>$. Then by the definition of order of a cyclic
  group, we have
$$a^n = e$$ But $m | n \implies n = mn'$ for some $m \in \mathbb{Z}$.
  Hence
$$a^{mn'} = e$$
and as a result, $\forall k \in \mathbb{Z}$ we have
$$a^{mn'k} = e \implies \left(a^{n'k}\right)^m = e$$
That is to say, the number of solutions to $x^m = e$ is equal to the
  number of multiples of $n'$ less than or equal to $n$, because $a^n = e$ can be factored out of any power of $a^h$ for $h > n$ to give a repeated solution. There are
  $\frac{n}{n'} = \frac{n}{\frac{n}{m}} = m$ of these.

Are all my steps, especially the last one, valid? How can I improve my answer if necessary?
 A: $G$ is a cyclic group of order n. Then the equation $x^n=e$ have exactly 
$n$ solution in $G$. Now if $m|n$ imply there exist a unique cyclic subgroup of order $m$ in $G$ hence there should be exactly $m$ solution for the equation $x^m=e$ in $G$. 
A: I think your proof is OK...as far it goes. Let me try to paraphrase it:
Let $d = \dfrac{n}{m}$. This is a (positive) integer, since $m|n$.
As you have shown:
$x = e, a^d, a^{2d},\dots , a^{(m-1)d}$ are $m$ distinct elements of $G = \langle a\rangle$ that all satisfy $x^m = e$ (these are all different, since for $0 \leq k < m \in \Bbb Z^+$, we have $0 \leq kd < md = n$).
However, you have not shown that these are the only solutions.
So suppose we have $x^m = e$, where $x = a^t$, and $d\not\mid t$.
We can, of course, write $t = qd + r$, where $0 < r < d$.
But $e = x^m = (a^t)^m = (a^{qd+r})^m = (a^{qdm})(a^{rm}) = (a^{dm})^q(a^{rm})$
$= (a^n)^q(a^{rm}) = e^q(a^{rm}) = a^{rm} = (a^r)^m$.
Thus if such a $t$ exists, we can assume $t < d$ (by using $r$ instead).
But $e = (a^r)^m = a^{rm}$, means that $a$ has order less than $rm < dm = n$, which is absurd. So no such $t$ (and thus $r$) can exist, we have accounted above for any possible element $x \in\langle a\rangle$ for which $x^m = e$.
