Power of 2001

Since $a_{n+1}= \frac {1}{1-a_{n}}$ for $n \geq 1$ , we get

$a_{3}= \frac {1}{1-a_{2}}= \frac {1}{1- \frac {1}{1-a_{1}}} = 1- \frac {1}{a_{1}}$

Also, it is given that $a_{3}=a_{1}$ , we get $a_{1}^{2}-a_{1}+1=0$ . Solving this equation, we get $a_{1}= \frac {1 \pm i \sqrt{3}}{2}$ .

Clearly, $a_{5}=1- \frac {1}{a_{3}}=1- \frac {1}{a_{1}}=a_{1}$ . Proceeding similarly, or by using mathematical induction, we can show that $a_{n}=a_{1}= \frac {1 \pm i \sqrt{3}}{2}=cos( \frac { \pi}{3}) \pm i\,sin( \frac { \pi}{3})$ for all odd values of $n$ .

Hence, $(a_{2001})^{2001}=(a_{1})^{2001}=(cos( \frac { \pi}{3}) \pm i\,sin( \frac { \pi}{3}))^{2001}$ .

Using De Moivre's Theorem, we get

$(a_{2001})^{2001}=cos( \frac {2001 \pi}{3}) \pm i\,sin( \frac {2001 \pi}{3})=cos(667 \pi) \pm i\,sin(667 \pi)=-1$