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 very small arc $a1b$ of the unit circle so that $a$ is above the real line, and $b$ below. Since 1 is a simple root of $P$, $h(z):=P(z)/z^n$ is a diffeomorphism of a neighborhood of 1 and neighborhood of 0. 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 $h(b)$. Let $\gamma_0$ be an $h$-preimage of a segment $[h(b),h(a)]$ between $h(b)$ and $h(a)$, it is a small curve near 1 joining $b$ and $a$. Let $\Gamma$ be a union of the large arc $ab$ of the unit circle and $\gamma_0$. It is a simple contour: for $z$ on the unit circle which is close to 1 but, say, higher then $a$, the directions of $h(z)-h(a),h(a)-h(b)$ are almost the same, so $h(z)$ can not belong to $[h(b),h(a)]$, and for $z$ which is far from 1 the value $h(z)$ is well away from 0, so again $h(z)\notin [h(b),h(a)]$. Then, 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 (it depends on whether 1 is inside $\Gamma$ or not). That's what you need.