A cyclic group is a group that can be generated by repeatedly applying the group operation to an element of the group called the "generator".
I noticed that, at least for the multiplicative group of integers modulo $5$, the equations $x = 2^y \pmod{5}$ and $x = 3^y \pmod{5}$, where $x$ is any element of said group, both have infinite solutions. $2$ and $3$ are generators of this group.
Ascending powers of either generator modulo 5 produce a unique, endless pattern. For example:
$$2^1\pmod{5} = 2$$
$$2^2\pmod{5}= 4$$
$$2^3\pmod{5} = 3$$
$$2^4\pmod{5} = 1$$
$$2^5\pmod{5} = 2$$
$$2^6\pmod{5}= 4$$
$$2^7\pmod{5}= 3$$
The repeating pattern contains all the elements of the group.
Is there a proof that any cyclic group generator produces an endlessly repeating pattern containing all elements of its group?
EDIT: I just realized what I'm asking is answered here: Why does a number like $2^n\bmod 7$ have a repeating pattern?
Yes.
For all generators $a$ of a cyclic group $G$, for each $g\in G$, there exists an $n\in\Bbb N$ such that $a^n=g$ by definition of $G$ being cyclic; that includes letting $g$ be another generator. For finite groups, it follows from the Pigeonhole Principle. For the only infinite cyclic group, $(\Bbb Z, +)$, up to isomorphism, the generator is either $1$ or $-1$, each of which generates each element as a multiple.