I'm mostly looking for advice on how to say, how to write this, how to prove it. I think it probably dispenses a proof. I'm writing this to a small number-theory audience, so I don't think they need much of anything here, but perhaps I could be in need of some help myself.
The prose. Take any iterated linear polynomial of integer coefficients such as one from the family
$$f(x) = ax \bmod{N}$$
with $1 < a \bmod{N}$ for some fixed natural number $a$ coprime with $N$. It is easy to see that any such polynomial from this family has closed-form formula $g(i) = a^i \bmod{N}$. Since $g$ generates a subgroup of $(Z/NZ, *)$, then we know that the length of the cycle of $g$ is the order of $a$ modulo $N$.
Should I prove this more formally? Because the closed-form formula is trivially obtained, we can also easily see that $a$ generates the set $\{a, a^2, a^3, ..., a^{r - 1}, a^r = 1\}$. Therefore, $f$ generates a cyclic subgroup of $(Z/ZN, *)$ with generator $a$.
Thank you for any thoughts.
An example. Choose $a = 2$ and $N = 15$ so that $f(x) = 2x \bmod{15}$. Let $x_0 = 1$. If we iterate $f$, we get the values $1, 2, 4, 8, 1, ...$. So, clearly, the closed-form formula for $f$ is $g(i) = 2^i \bmod{15}$, which suggests that $2$ is a generator for the subgroup {1, 2, 4, 8} of $(Z/15Z, *) = \{1, 2, 4, 6, 7, 8, 11, 13, 14\}$.