Roots of $x^n-x^{n-1}-\cdots-x-1$

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?