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First approach (Evan Chen) We introduce the following notations: $$\begin{array}{rl}P(x)& =c_n{x}^n+c_{n-1}{x}^{n-1}+\cdots +c_{1}x+c_0\\ & =c_n(x+{\alpha }_1)\dots (x+{\alpha }_n)\\ P(x)-1& =c_n(x+{\beta }_1)\dots (x+{\beta }_n)\end{array}$$ By taking conjugates, $$⟹\left(x+\frac{1}{{\alpha }_1}\right)\cdots \left(x+\frac{1}{{\alpha }_n}\right)=\left(x+\frac{1}{{\beta }_1}\right)\cdots \left(x+\frac{1}{{\beta }_n}\right)+{\left(\overline{c_n}\right)}^{-1}(♠)$$ The equation (♠) is the main player: Claim — We have $c_k=0$ for all $k=1,\dots ,n-1$. Proof. By comparing coefficients of $x^k$ in (♠) we obtain $$\frac{c_{n-k}}{{\prod }_i{\alpha }_i}=\frac{c_{n-k}}{{\prod }_i{\beta }_i}$$ but ${\prod }_i{\alpha }_i-{\prod }_i{\beta }_i=\frac{1}{c_n}\ne 0$. Hence $c_k=0$. $\square$ It follows that $P(x)$ must be of the form $P(x)=\lambda x^n-\mu$, so that $P(x)=\lambda x^n-(\mu +1)$. This requires $|\mu |=|\mu +1|=|\lambda |$ which is equivalent to the stated part.

Second approach (from the author) We let $A=P$ and $B=P-1$ to make the notation more symmetric. We will as before show that $A$ and $B$ have all coefficients equal to zero other than the leading and constant coefficient; the finish is the same. First, we rule out double roots. Claim — Neither $A$ nor $B$ have double roots. Proof. Suppose that $b$ is a double root of $B$. By differentiating, we obtain $A^{\prime }=B^{\prime }$, so $A^{\prime }(b)=0$. However, by Gauss-Lucas, this forces $A(b)=0$, contradiction. $\square$

Claim (Main claim) — For any $i$ and $j$, $\frac{a_i}{a_j}$ is a power of $\omega$. Proof. Note that $$\frac{a_i-b_1}{a_j-b_1}\cdots \frac{a_i-b_n}{a_j-b_n}=\frac{B(a_i)}{B(a_j)}=\frac{A(a_i)-1}{A(a_j)-1}=\frac{0-1}{0-1}=1.$$ Since the points $A_i$, $A_j$, $B_k$ all lie on the unit circle, interpreting the left-hand side geometrically gives $$\angle A_i{B}_1{A}_j+\cdots +\angle A_i{B}_n{A}_j=0⟹n\widehat{A_i{A}_j}=0,$$ where angles are directed modulo $180^{\circ }$ and arcs are directed modulo $360^{\circ }$. This implies that $\frac{a_i}{a_j}$ is a power of $\omega$. $\square$ Now the finish is easy: since $a_1,\dots ,a_n$ are all different, they must be $a_1{\omega }^0,\dots ,a_1{\omega }^{n-1}$ in some order; this shows that $A$ is a multiple of $x^n-a_1^n$, as needed.