$\text{ If }{\mathrm{a}}_{\mathrm{n}+1}=\frac{1}{1-{\mathrm{a}}_{\mathrm{n}}}\text{ for }\mathrm{n}\ge 1\text{ and }{\mathrm{a}}_3={\mathrm{a}}_1\text{ then }\left|{\left({\mathrm{a}}_{2001}\right)}^{2001}\right|\text{ is }$

$\begin{array}{l}{\mathrm{a}}_2=\frac{1}{1-{\mathrm{a}}_1},{\mathrm{a}}_3=\frac{1}{1-{\mathrm{a}}_2}=\frac{1}{1-\frac{1}{1-{\mathrm{a}}_1}}=\frac{1-{\mathrm{a}}_1}{-{\mathrm{a}}_1}\\ \text{ Since }{\mathrm{a}}_3={\mathrm{a}}_1\to -{\mathrm{a}}_1^2=1-{\mathrm{a}}_1\\ \Rightarrow {\mathrm{a}}_1^2-{\mathrm{a}}_1+1=0\Rightarrow {\mathrm{a}}_1=-{\mathrm{w}}_{,}-{\mathrm{w}}^2\\ {\mathrm{a}}_5=\frac{1}{1-{\mathrm{a}}_4}=\frac{1}{1-{\displaystyle \frac{1}{1-a_3}}}=\frac{{\mathrm{a}}_3-1}{{\mathrm{a}}_3}\end{array}$

$\begin{array}{l}\therefore \text{ continuing in their way, we obtain }\\ {\mathrm{a}}_1={\mathrm{a}}_3={\mathrm{a}}_5=\dots ={\mathrm{a}}_{2001}\Rightarrow {\left({\mathrm{a}}_{2001}\right)}^{2001}=(-\mathrm{w}{)}^{2001}=-1\\ \Rightarrow \left|{\left({\mathrm{a}}_{2001}\right)}^{2001}\right|=1\end{array}$