How do you find nth term rule for 1,3,9,27,...?

The n-th term of this sequence appears to be ${3}^{n - 1}$, $n \ge 1$.
These are powers of $3$ ordered from ${3}^{0} = 1$ to ${3}^{a}$ (for an integer $a \ge 1$). However, the first convenient value for $n$ is $1$, not $0$ (imagine saying the 0th term of a sequence). Because of that, since the first term is actually ${3}^{0}$, we need to start from the first term ($n = 1$) being
${3}^{1 - 1}$. The next is ${3}^{2 - 1}$, ${3}^{3 - 1}$ ... ${3}^{n - 1}$.