Easily gotten: $a$ is even and $n=2^k$ for some $k \geq 0$. Thus, $a^n-1 = a^{2^k} -1 =(a-1)\prod_{r=0}^{k-1} \left(a^{2^r}+1\right)$. The terms $a^{2^u}+1$ and $a^{2^v}+1$ are relatively prime if $u \neq v$. In addition, $a^{2^r}+1$ is an odd number greater than $1$ for all $r=0,1,2,\ldots,k-1$. We conclude that $\prod_{r=0}^{k-1} \left(a^{2^r}+1\right)$ has at least $k$ distinct prime divisors. That is, \[d\left(a^n-1\right) \geq 2^k = n\,.\]