How to solve this algebraic problem about remainder of polynomials? Question :

My approach :
Now as I had to obtain a remainder of $\frac{(x-1)^{2017}}{x^2 - x +1}$
So, I could write this as $\frac{(x-1)^{2017}}{(x - 1)^2 + x}$
now I substitute $t = (x-1)$, so $\frac{(x-1)^{2017}}{(x - 1)^2 + x}$ could be written as $\frac{t^{2017}}{t^{2} + t + 1}$
From here we can easily obtain the remainder by plain division itself, so $P^{1}(x) = -(t+1)t^{2015} = -x(x-1)^{2015}$
Since we need $P^{2017}(2016)$, using the above obtained $P^{1}(x)$, I first obtained $P^{1}(2016) = -2016(2015)^{2015}$.
Similarly $P^{2}(x) = -P^{1}(x)(P^{1}(x) - 1)^{2015}$ by question definition, but there's no further way to simplify it, and also $P^{n}(x)$ gets bigger and bigger as $n$ increases.
So I genuinely feel this approach is wrong way, so could you help me out upon this question ?

$p(x) = q(x) d(x) + P^1(x)$
$P^{1}(x)$ is at most a 1 degree polynomial
If $\omega$ is a root of $d(x).$
$p(\omega) = P^1(\omega)$
From this we find $P^{1}(\omega_1), P^{1}(\omega_2)$ and the line between them.
$\omega_1, \omega_2 = \frac 12 + i \frac {\sqrt 3}{2}, \frac 12 - i \frac {\sqrt 3}{2}$ 
$p(\omega_1) = $$(-\frac 12 + i \frac {\sqrt 3}{2})^{2017}\\
e^{\frac {2017\cdot2 \pi}{3}}\\
-\frac 12 + i \frac {\sqrt 3}{2}\\
\omega_1 - 1$
$p(\omega_2) = \omega_2 - 1$
$P^1(x) = x-1$
$P^2(x) =$$ P^1(P^1(x))\\
x-2$
$P^{k}(x) =x-k$
$P^{2017}(2016)=-1$
