If $a_{n+1}=\frac{1}{1-a_n}$ for $n\ge 1$ and $a_3=a_1,$ then ${\left(a_{2022}\right)}^{2022}$ equal 

We have $a_2=\frac{1}{1-a_1}$ and $a_3=\frac{1}{1-a_2}=\frac{1}{1-1/\left(1-a_1\right)}=\frac{1-a_1}{-a_1}$

Since $a_3=a_1,$ we get $-a_1^2=1-a_1$

$\Rightarrow a_1^2-a_1+1=0\Rightarrow a_1=-\omega$ or $-{\omega }^2$

We have $\begin{array}{r}{a}_5=\frac{1}{1-a_4}=\frac{1}{1-1/\left(1-a_3\right)}=\frac{1-a_3}{-a_3}=\frac{1-a_1}{-a_1}\\ =a_3=a_1\end{array}$

Continuing in this way we obtain

$\begin{array}{r}{a}_1=a_3=a_5=a_7=a_9=a_{11}=\dots =a_{2022}\\ \text{ Thus, }{\left(a_{2022}\right)}^{2022}=(-\omega {)}^{2022}\text{ or }{\left(-{\omega }^2\right)}^{2022}=1\end{array}$