Show that for any positive integer $k$, the order of $a^k$ is $\frac{d}{gcd(k,d)}$.
We have $a$ relatively prime to n, with the order of $a\ mod\ n$ denoted by $d$.
Specifically, the highlighted bit has me confused. How do we know $gcd(\frac{d}{gcd(k,d)}, \frac{k}{gcd(k,d)}) = 1$?
By the fundamental theorem of arithmetic, we can write $d$ as $d = p_1p_2\dots p_r$ and $k$ as $k = q_1q_2\dots q_s$ where the $p_i$'s and $q_j$'s are all primes, uniquely determined up to reordering and sign. We can assume that we ordered them so that the first $t$ of them are the same, that is, $p_1 = q_1, \dots, p_t = q_t$ (of course, $t$ could be $0$ if none of them are the same). Then the greatest common divisor is just $p_1\dots p_t = q_1\dots q_t$ Now, when we divide through by this, the resulting numbers don't have any primes in common, so their gcd has to be 1.
