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 (a.k.a. [PV number](https://en.wikipedia.org/wiki/Pisot%E2%80%93Vijayaraghavan_number)), i.e., the other roots are in the open disk $\{z\in \mathbb{C}: |z|<1\}$. 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 Rouché'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? 



Roots of $x^n-x^{n-1}-\cdots-x-1$
Let $r\in\mathbb{R}$ be slightly larger than $1$. Then, on the circle $|z|=r$, we have
$$|z^{n+1}+1|\leq r^{n+1}+1<2r^n=|2z^n|,\qquad |z|=r.$$
By Rouché's theorem, it follows that $P(z):=z^{n+1}-2z^n+1$ has the same number of zeros in the open disk $|z|<r$ as $2z^n$ does (zeros are counted with multiplicity). Therefore, $P(z)$ has $n$ roots in the open disk $|z|<r$. This is true for any $r\in\mathbb{R}$ slightly above $1$, hence $P(z)$ has $n$ roots in the closed disk $|z|\leq 1$. One of these roots is $z=1$, and we claim that there is no other root on the circle $|z|=1$. Once we prove this, it follows that $P(z)$ has $n-1$ roots in the open disk $|z|<1$, and the same is true of
$$P(z)/(z-1)=z^n-z^{n-1}-\dots-z-1.$$


So let $|z|=1$ and assume that $P(z)=0$. Then $|z^{n+1}+1|=|2z^n|=2$. By the triangle inequality, this is only possible when $z^{n+1}=1$. However, in that case, $2z^n=z^{n+1}+1=2$, hence $z^n=1$ as well, and we conclude that $z=z^{n+1}/z^n=1/1=1$. Done.


