This is the proof in my textbook:
Proof: Since $d\mid n$, then $n = dk$, for some $k$. Let $a$ be a generator of $G$ and consider $a^k$. Clearly $(a^k)^d= a^{kd} = e$. If the order of $a^k$ were smaller than $d$, then the order of $a$ would be smaller that $n$, contradiction. Therefore $a^k$ has order $d$ and so generates a subgroup of order $d$
I dont understand this proof. Why is this a contradiction. Can someone give me an intuition of why this theorem should be true and how this proof works?
Nevermind I answered it myself. Here is the answer in case someone in the future has the same problem as me and can't figure it out.
Consider the element $a^k$ in $G$. Notice that $(a^k)^d = a^{(kd)} = a^n = e$, where $e$ is the identity element of $G$. This means that the order of $a^k$ is at most $d$.
We want to show that the order of $a^k$ is exactly $d$. Suppose that the order of $a^k$ is less than $d$. Then there exists a positive integer $ m < d$ such that $(a^k)^m = e$. This means that $a^{(km)} = e$, which implies that the order of $a$ is at most $km$. But since $n = dk$ and $m < d$, we have $km < kd = n$, which contradicts the fact that the order of $a$ is $n$.