For a group $G$ and element $a$ in $G$, the order of $a$, denoted $|a|,$ is the smallest positive integer $n$ so that applying the group operation $n$ times to $a$ yields the identity element. Let $G$ be a group and let $a\in G$ with $|a| = n.$ Prove that for each $k\in\mathbb{Z},$ the order of $a^k$ is a positive divisor of $n$ and for each positive divisor $d | n,$ the number of elements in $\langle a \rangle$ of order $d$ equals $\phi(d),$ the Euler totient function evaluated at $d$.
Prove that for a finite group $G$ and each $d \in \mathbb{Z}^+$, the number of elements in $G$ of order $d$ equals $\phi(d)$ times the number of cyclic subgroups of $G$ of order $d$.
The second claim seems to follow easily from the first in my opinion. To prove the first one, I know how to show that the order of $a^k$ is $\frac{n}{\gcd (k,n)}$. Indeed, $(a^k)^{(n/\gcd(k,n))} = e$ where $e$ is the identity element, and for any $ i < n/\gcd(k,n), (a^k)^i\neq e$ as $|a| = n.$ Then suppose $d | n$ is a positive divisor of $n$. There is a unique subgroup of $G, \langle a^{n/d}\rangle =: \langle b\rangle$ that has order $d.$
Why is it the case that every element of order $d$ generates $\langle b\rangle? \tag{*}$
Given this, take an element $a^k$ that generates $\langle b\rangle$. Then since $|\langle b\rangle| = d,\langle a^k\rangle = \langle b\rangle\Leftrightarrow gcd(k,d) = 1,$ and so every element of order $d$ is coprime to $d$, which implies that the number of elements in $\langle a\rangle$ of order $d$ is $\phi(d).$ Supposing the justification tof $(*)$ is complete, this proves the first part.
For the second part, let $G$ be a finite group and $d\in \mathbb{Z}^+.$ By $(*),$ every element of order $d$ generates the cycles of order $d$. The number of elements that generate a cycle of order $d$ is $\phi(d)$ as shown above. Since this applies for each cycle of order $d$, the number of elements of order $d$ equals $\phi(d)$ times the number of cycles of order $d$.
Is my second justification correct, assuming (*) holds?