It is easy to see that $ f(x)=x^n-x^{n-1}-\cdots-x-1$ has only one positive root $ \alpha$ which lies in the interval $ (1,2)$ . But it is claimed that this root is a Pisot number, i.e., the other roots are in the open $ \{z\in \mathbb{C}: |z|<1\}$ . I have tried the following, but I failed. I considered $ $ P(x)=(1-x)f(x)=x^{n+1}-2x^n+1=x^n(x-2)+1$ $ and then I tried to use Rouche’s theorem, by picking $ g(x)=-x^n(x-2)$ and trying to show that $ 1<|g(z)|+|P(z)|$ for $ |z|=1$ . Proving this implies that $ P(x)$ has $ n$ roots in $ \{z\in \mathbb{C}: |z|<1\}$ which demonstrates the claim. But this inequality fails at $ z=1$ . Can you give me an idea how one can prove this claim?