A sequence $(a_n)$ is defined as follows:
\[a_{i + 1} = \frac{1}{1 - a_i}\]for $i \ge 1.$  If $a_3 = a_1,$ compute $(a_9)^9.$
First, if $a_3 = a_1,$ then
\[a_1 = a_3 = a_5 = a_7 = a_9,\]so $(a_9)^9 = (a_1)^9.$

We have that
\begin{align*}
a_2 &= \frac{1}{1 - a_1}, \\
a_3 &= \frac{1}{1 - a_2} = \frac{1}{1 - \frac{1}{1 - a_1}} = \frac{1 - a_1}{1 - a_1 - 1} = \frac{1 - a_1}{-a_1}.
\end{align*}Then
\[\frac{1 - a_1}{-a_1} = a_1,\]so $1 - a_1 = -a_1^2.$  Then $a_1^2 - a_1 + 1 = 0.$  Multiplying both sides by $a_1 + 1,$ we get
\[(a_1 + 1)(a_1 ^2 - a_1 + 1) = 0,\]so $a_1^3 + 1 = 0.$  Then $a_1^3 = -1,$ so $a_1^9 = (-1)^3 = \boxed{-1}.$