Find the root of the polynomial?

Consider the root of the polynomial $p(x) = x^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+\dots+a_1x -1$. Suppose that $p(x)$ has no roots in the open unit disc in a complex plane and $p(-1)=0$. Show that $p(1)= 0$.

This is not true if the $a_k$ aren't assumed real, e.g. $p(x) = (x+1)(x - i)^2$.

Let's assume now the coefficients are real. The roots are all unit complex numbers because their product has modulus $1$ and they are all of modulus $\geqslant 1$.

If $1$ was not a root of $p$, then $p$ would have an expression of the form $$p(x) = (x + 1)^{m}(x-\alpha_1)^{m_1}(x-\overline{\alpha_1})^{m_1} \dots (x-\alpha_k)^{m_k}(x-\overline{\alpha_k})^{m_k}$$ where $\alpha_1, \dots \alpha_k$ are the non real roots of $p$ (they are unit complex numbers) and $m, m_1, \dots, m_k$ are positive integers that add up to $n$.

But this polynomial satisfies $p(0) = 1$, contrary to the hypothesis.