Find all $n\in\mathbb{N}$ such that the nonzero roots of $(z+1)^n-z^n-1$ are all on the unit circle. I try to use Mathematica to find such $n$, and I believe that only $n=2,3,4,5,6,7$ meet the requirement but I don't know how to prove that for $n\ge 8$, there is always some root lying outside the unit ball by using the basic complex analysis knowledge. 
In fact, I do have a dirty way to deal with this problem, which only involves the property of polynomial. But I don't think it is an efficient and elegant solution. 
 A: It is clear that the result is true for $n=2$. Now we consider $n\ge 3$.
Note that $(z+1)^n-z^n-1=\sum_{k=0}^n\binom{n}{k} z^k-z^n-1 = \sum_{k=1}^{n-1}\binom{n}{k} z^k= nz\sum_{k=1}^{n-1}\frac{\binom{n}{k}}{n}z^{k-1}$. Suppose that $\sum_{k=1}^{n-1}\frac{\binom{n}{k}}{n}z^{k-1}=\prod_{k=1}^{n-2}(z-z_k)$. Then we have $\sum_{k=1}^{n-2}z_k=-\frac{\binom{n}{n-2}}{n}=-\frac{n-2}{2}$ and $\sum_{j\ne l}z_jz_l=\frac{\binom{n}{n-3}}{n}=\frac{(n-1)(n-2)}{6}$. Then
$$\sum_{k=1}^{n-2}z_k^2 = \left(\sum_{k=1}^{n-2}z_k\right)^2-2\sum_{j\ne l}z_jz_l = \frac{(n-1)^2}{4}-\frac{(n-1)(n-2)}{3}= -\frac{n^2-6n+5}{12}.$$
Write $z_k=x_k+iy_k$, where $x_k,y_k\in\mathbb{R}$ for $k=1,\cdots, n-1$. Note that for each root $z_k$, $\overline{z_k}$ is also a root and $z_k^2+\overline{z_k}^2 = x_k^2-y_k^2+2ix_ky_k+x_k^2-y_k^2-2ix_ky_k=2(x_k^2-y_k^2)$, so 
$$\sum_{k=1}^{n-2}z_k^2=\sum_{k=1}^{n-2}(x_k^2-y_k^2)=-\frac{n^2-6n+5}{12}.$$
Suppose that all the nonzero roots lying on the unit circle, then
$$\sum_{k=1}^{n-2}|z_k|^2 = n-2=\sum_{k=1}^{n-2}(x_k^2+y_k^2),$$
with the above equation, we have
$$2\sum_{k=1}^{n-2}x_k^2 = n-2-\frac{n^2-6n+5}{12}=-\frac{n^2-18n+29}{12}\ge 0,$$
which follows that $n\le 9+2\sqrt{13}\approx 16.2$. 
By using the Wolfram Mathematica, for $n=2,3,4,5,6,7$, the nonzero roots of $(z+1)^n-z^n-1=0$ lying on the unit circle and for $8\le n\le 16$, there is some root not lying on the unit circle.
A: I come up with another solution:
Gauss-Lucas Theorem: let $p(z)$ be the polynomial with coefficient in $\mathbb{C}$, then zeros of $p'(z)$ lie in the convex hull of zeros of $p(z)$.
Let $n\ge 3$. Here, $p(z)=(z+1)^n-z^n-1$ and $p'(z)=n((z+1)^{n-1}-z^{n-1})$. It is clear that $p'(0)\ne 0$ and hence $p'(z)=nz^{n-1}\cdot(\frac{z+1}{z})^{n-1}$, which follows that the zeros of $p'(z)$ are $\frac{z+1}{z}=e^{\frac{2\pi i k}{n-1}}$, $k=1,\cdots,n-2$, so $z=\frac{1}{e^{\frac{2\pi i k}{n-1}}-1}$. 
Note that the zeros of $p(z)$ lie on the unit circle, by Gauss-Lucas Theorem, the zeros of $p'(z)$ lie in the unit ball. For each zero of $p'(z)$,
$$\left|\frac{1}{e^{\frac{2\pi i k}{n-1}}-1}\right|=\frac{1}{2-2\cos(\frac{2\pi k}{n-1})}\le 1,$$
which follows that $\cos(\frac{2\pi k}{n-1})\le\frac{1}{2}$, i.e., $\frac{\pi}{3}\le \frac{2\pi k}{n-1}\le \frac{5\pi}{3}$ for all $k=1,\cdots,n-2$. Then we have $\frac{\pi}{3}\le \frac{2}{n-1}$ and $\frac{2(n-2)}{n-1}\le \frac{5}{3}$, so $n\le 7$.
It is easy to verify that the result holds for $n=2,3,4,5,6,7$. 
