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} $.