<s>[INST] 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. 
 [/INST] 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.
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