Hint: this follows from the periodicity of recurrences on finite rings. The power sequence $f_k = a^k\,$ satisfies the recurrence $\,f_{k+1} = a f_k,\ f_0 = 1.\,$ Since it take values in a finite ring $\, R = \Bbb Z_{30}\,$ the Pigeonhole Principle implies it eventually repeats: say $\,\color{#c00}{f_{j+n} = f_j}\,$ so induction using the recurrence shows it continues to repeat in-step for larger indices, i.e.
$$\begin{align} \color{#0a0}{f_{1+j+n}} &= a \color{#c00}{f_{0+j+n}} = a \color{#c00}{f_{0+j}} = \color{#0a0}{f_{1+j}},\ \ \ {\rm i.e.}\ \ \ \,a \ \left[ f_{n+j} = f_n\right]\rightarrow f_{1+j+n} = f_{1+j}\\
f_{2+j+n} &= a \color{#0a0}{f_{1+j+n}} = a \color{#0a0}{f_{1+j}} = f_{2+j},\ \ \ {\rm i.e.}\ \ \ a^2 \left[ f_{n+j} = f_n\right]\rightarrow f_{2+j+n} = f_{2+j}\\
&\ \ \,\vdots\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\ \ \vdots\\
&\phantom{a \color{#0a0}{f_{k-1+j+n}} = a \color{#0a0}{f_{1+j}} =
f_{2+j},\ \ \ \ \ }{\rm i.e.}\ \ \ a^k \left[ f_{n+j} = f_n\right]\rightarrow f_{k+j+n} = f_{k+j}
\end{align}\qquad$$
Hence, $\, f_K = f_k\,$ if $K\equiv k\pmod{\! n}$ and $\,K,k\ge j,\,$ i.e. once both indices are $\ge j\, $ we enter a cycle of length $\,n,\,$ i.e. $\rm\color{#0af}{MOR}$ = mod order reduction eventually applies. This holds even if $\,j\,$ and $\,n\,$ are not the minimal possible values. When we choose them minimal - so $n$ is the order of the cycle - then we get a converse - just like in $\rm\color{#0af}{MOR}$.
Remark: $\! $ oh vs. rho orbits: $\!$ permutation orbits are cycles, i.e. o-shaped vs. $\rho$-shaped
When $\,a\,$ is coprime to the modulus $\,n\,$ then the shift map $f_n \to f_{n+1} = a f_n\,$ is invertible, so being an invertible map on a finite set it is a permutation, whose orbits are purely periodic, i.e. cycles, i.e. o-shaped (vs. generally having a preperiodic part, i.e. $\rho$-shaped, i.e. $\,j> 0).\,$ This simple general fact about such periodicity is often overlooked, resulting in reinventing the wheel (cycle).
Generally the same argument works over any finite commutative ring $R$ when we have a (nonlinear) recurrence of order $k$ of form $\,f_{n+k} = g(f_{n+k-1},\ldots,f_0)$ for $\,g\,$ a polynomial over $R$, i.e. where the next value is a polynomial function of the prior $k$ values. As above it is eventually periodic by $R$ is finite so there are only finitely many sequences of $k$ values from $R$ so they must eventually repeat as a subsequence in the sequence $\,f_n,\,$ hence the values of $f_i$ repeat in-step after these matching points, by induction - as above. If the recurrence is linear then we can represent the shift map as a matrix $A$ and then the repetition occurs via scaling by powers of $A$. In particular $\,f_n\,$ can be computed efficiently by repeated squaring of $A$, e.g. as here for fast computation of Fibonacci numbers.
