$30^n=3^n\cdot 10^n$, but since you are interested in the last non-zero digit, you need not consider $10^n$ (since $30^n$ is just $3^n$ with $n$ following $0$s).
Besides, $3^n$ may be recursively defined as
$$
\begin{cases}
3^0 = 1 \\
3^n = 3\cdot3^{n-1} \quad n\geq 1
\end{cases}
$$
It is trivial that the last digit of $3^n$ is never $0$, and that it depends only on that of $3^{n-1}$; to be more precise, if the last digit of $3^{n-1}$ is $d$, then the last digit of $3^n$ is the last digit of $3d$. From the list of computations you provided, by a trivial induction it follows that the only possibilities for the last digit of $3^n$ are four, namely 1,3,9,7, and these occur recursively in this order.