jmc

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 $$\begin{array}{rl}{a}_2& =\frac{1}{1-a_1},\\ a_3& =\frac{1}{1-a_2}=\frac{1}{1-\frac{1}{1-a_1}}=\frac{1-a_1}{1-a_1-1}=\frac{1-a_1}{-a_1}.\end{array}$$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 $a_1^9=(-1{)}^3=\boxed{-1}.$