Note that a straightforward induction reveals \[ a_i = mn^i + \frac{n^i-1}{n-1},\quad 1\le i\le m-1. \]Now, assume first $m$ is a prime. As $(m,n)=1$, it follows by Fermat that $m\mid n^{m-1}-1$. Further, $m\nmid n-1$ as $(m,n-1)=1$. So, $m\mid a_{m-1}$. But as $a_{m-1}>m$, it must be composite. Next, assume $m$ is composite, let $p$ be a prime dividing $m$. Note that $p-1\in\{1,\dots,m-1\}$. Further, $(p,n)=1$ implies via Fermat that $p\mid n^{p-1}-1$, whereas $(m,n-1)=1$ implies $p\nmid n-1$. So, $p\mid a_{p-1}$. Clearly $a_{p-1}>p$, which shows $a_{p-1}$ is composite.