For $P(x)=(1-x)f(x)=(x^{n+1}+1)-2x^n$ you may apply a version of Rouché's theorem or argument principle: 

choose a small arc $a1b$ of the unit circle so that 1 is its midpoint, and consider a circle $\gamma$ centered in 1 passing through $a$ and $b$. Denote by $\gamma_1$ and $\gamma_2$ two arcs of $\gamma$ between $a$ and $b$. Since 1 is a simple root of $P$, $h(z):=P(z)/z^n$ makes one full rotation around 0 when $z$ goes along $\gamma$. When $z$ goes along a large arc $ab$ of the unit circle, $h(x)=P(x)/x^n=-2+(x+1/x^n)$ stays in the left half-plane and goes from $h(a)$ to $h(b)$. Thus, if we add either $h(\gamma_1)$ or $h(\gamma_2)$ to this path, we get a closed contour of the form $h(\Gamma)$ with 0 full rotations around 0, where $\Gamma$ is a union of arc $ab$ of the unit circle and either $\gamma_1$ or $\gamma_2$. Thus, by the argument principle, $x^n$ and $x^nh(x)=P(x)$ have equally many roots inside $\Gamma$. Therefore $P$ has $n$ roots inside $\Gamma$, and $f$ has either $n-1$ or $n$ roots inside the unit circle. That's what you need.