Problem: Let $a$ be an element of order $n$ in a group and let $k$ be a positive integer. Then $\langle a^k \rangle = \langle a^{gcd(n,k)}\rangle$.
My proof: Let $d=gcd(n,k)$, so there exist $r ∈ ℤ $ s.t $k=dr$. Suppose $a^{ki} \in \langle a^k \rangle$. We have $a^{ki} = a^{dri}= (a^d)^{ri} ∈ \langle a^d \rangle$. Thus $\langle a^k \rangle \subseteq \langle a^d \rangle $. For other inclusion, assume $a^{dj} ∈ \langle a^d \rangle$. We see $a^{dj} = a^{jpn+jqk}= (a^k)^{jq} ∈ \langle a^k \rangle$ for some $p,q \in \mathbb Z$. So $\langle a^d \rangle \subseteq \langle a^k \rangle$.
Proof in the book: Let $d=gcd(n,k)$ and let $k=dr$. Since $a^k = (a^d)^r$, we have by closure that $\langle a^k \rangle ⊆ \langle a^d \rangle$. By gcd theorem there are integers $s $ ans $t$ s.t $d=na+tk$. So $a^d= a^{ns+tk}=a^{ns}a^{tk} = (a^k)^t \in \langle a^k \rangle$ . This proves $\langle a^d \rangle \subseteq \langle a^k \rangle$.
My confusion: Is my proof correct? Because I used arbitrary element to show each is subset of other. On the other hand author used arbitrary element.