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 (a.k.a. PV 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? 
 A: You may want to take a look at pages 155-156 of the problems and solutions section of the February 1989 issue of the American Mathematical Monthly.
You are going to find there two proofs of the irreducibility over $\mathbb{Q}$ of the polynomial $$p(x)=x^{n}-x^{n-1}-\cdots-1.$$
The first proof depends crucially on the fact that you wished to establish (which is settled therein via Rouché's theorem along the lines of the above proof by GH from MO).
The editorial comment you are to find on page 156 is noteworthy, too: in his paper "On algebraic equations with all but one root in the interior of the unit circle", Alfred T. Brauer proved that if $a_{1}, a_{2}, \ldots, a_{n}$ are integers with $a_{1} \geq a_{2} \geq \cdots \geq a_{n} > 0$, then the polynomial
$$x^{n}-a_{1}x^{n-1}-a_{2}x^{n-2}-\cdots-a_{n}$$ has one of its roots in the exterior of the unit circle and all the others in its interior.
Let me conclude this intervention by echoing, once again, Leo Sauvé's famed remark on the problems and solutions department of the Monthly: "it seems like all problems have once been published in the American Mathematical Monthly".
A: 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.
A: Let's prove the equivalent claim:

The only complex solutions to $2z=z^{n+1} +1$ with $|z|\le 1$ are $z={1\over\alpha}$ and $z=1$.

Remark: The number $\beta:={1\over\alpha}$ is the minimum fixed point of the increasing convex function $\displaystyle g( r):={1+r^{n+1}\over 2} $ on $\mathbb{R}_+$, and as such 
$0<g'(\beta)=(n+1)\beta^n<1$.
$\phantom{y}$
Proof of claim. Assume  $\zeta$ verifies $2\zeta=\zeta^{n+1} +1$ and $|\zeta|=r\le 1.$ 
Case 1: $r \le \beta$. Then subtracting $2\beta=\beta^{n+1} +1$ we get $$2|\zeta-\beta |= |\zeta^{n+1}-\beta^{n+1}|\le |\zeta-\beta|\sum_{k=0}^nr^k\beta^{n-k}\le|\zeta-\beta |(n+1)\beta^n.$$ But $0<(n+1)\beta^n=2g'(\beta)<2$.
So $\zeta=\beta$.
Case 2: $\beta< r \le1$. Then   $$2r=|\zeta^{n+1} +1| \le r^{n+1} +1 \le 2r,$$   so  $r^{n+1} +1 = 2r,$ and in the interval $(\beta,1]$ this forces $r=1$. Hence $|\zeta^{n+1} +1|=2$. So $\zeta^{n+1}=1$ because $|\zeta|\le1$. Then $2\zeta =\zeta^{n+1} +1=2$ and $\zeta=1.$ $\quad\square$
Passing to the reciprocal equation: the only complex solutions to $2x^n=x^{n+1}+1$ with $|x|\ge1$ are $x=\alpha$ and $x=1$, whence your thesis.
