Understanding a proof of Wedderburn in Topics in algebra
Suppose that $r$ is not prime. Then you can write $r$ as $r_1r_2$, where $r_1$ and $r_2$ are natural numbers greater than $1$. But then $(a^{r_1})^{r_2}=a^r\in Z$. So, if $b=a^{r_1}$, $b^{r_2}\in Z$. But $r_2<r$ and the author was supposing that $r$ is the smallest natural number such that $c^r\in Z$ for some $c\in D\setminus Z$. So, a contradiction is reached. It was a conclusion of the assumption that $r$ is a composite number. So, since $r$ cannot be $1$, it has to be a prime number.