How many zeros of $z^{17}+z-1$ lie in the region $|z|<1$? General problem:
Conjecture (B. Galo and Jack D'Aurizio) Let $p$ be a prime number then the number of zeros of $z^{p}+z-1$ that lie in the region $|z|<1$ is:
1)  $⌊\frac{p}{3}⌋-2$ (where $⌊*⌋$ is the lower integer function). If $ p≡5 (mod6)$ but $p\neq 5$.
2) $⌊\frac{p}{3}⌋$. If this result is odd and $p\not\equiv 5 (mod6)$
3) $⌊\frac{p}{3}⌋+1$. If $⌊\frac{p}{3}⌋$ is even number. 
I have to thanks Jack D'Aurizo for the "review" of my attempt and for sketching the proof.

Visualiztion
Start with
$$
f(z) = z^{17}+z-1
$$
The Re and Im components are shown next.

The $0$ contours, the line where
$$
\color{blue}{\text{Re } f= 0}, \qquad \color{red}{\text{Im } f= 0}
$$
are shown below against the unit circle. The roots where $f(z)=0$ are where the blue and red contours intersect. We see six roots within the unit disk.

Observation
While using the argument principle, a pattern emerges.
Given
$$
 g(k,z) = z^{k} - z + 1
$$
compute the series expansion for
$$
 \frac{g'(z)}{g(z)}
$$
The expansion coefficients are
$$
\begin{array}{lrrrrrrrrrrrrrrrrrrrr}
 k & const & w^1 & w^2 & w^3 & w^4 & w^5 & w^6 & w^7 & w^8 & w^9  & w^{10}\\\hline
 2 & -1 &  1 &  2 & \dots \\
 3 & -1 & -1 &  2 &  3 &  4 & \dots \\
 4 & -1 & -1 & -1 &  3 &  4 &  5 & 6 & \dots \\
 5 & -1 & -1 & -1 & -1 &  4 &  5 & 6 & 7 & 8 & \dots \\
 6 & -1 & -1 & -1 & -1 & -1 &  5 & 6 & 7 & 8 & \dots \\
 7 & -1 & -1 & -1 & -1 & -1 & -1 & 6 & 7 & 8 & 9 & 10 & 11 & 12 \dots \\
\end{array}
$$
