<s>[INST] Show that the following non-linear group of equations only have zero solution 
If
$$\left\{\begin{array}{c}
(\lambda_1 + 1)(\lambda_2 + 1) \cdots (\lambda_n + 1) = 1 \\
(\lambda_1^2 + 1)(\lambda_2^2 + 1) \cdots (\lambda_n^2 + 1) = 1 \\
\vdots \\
(\lambda_1^n + 1)(\lambda_2^n + 1) \cdots (\lambda_n^n + 1) = 1
\end{array}\right.$$
with $\lambda_1, \lambda_2, \ldots, \lambda_n \in \Bbb{C}$, then
$$\lambda_1 = \lambda_2 = \cdots = \lambda_n = 0.$$

Sorry I misunderstood the question before, actually it is stated as follows:

If
$$
\
(\lambda_1^k + 1)(\lambda_2^k + 1) \cdots (\lambda_n^k + 1) = 1\\
.$$
hold for all positive integers $k$, with $\lambda_1, \lambda_2, \ldots, \lambda_n \in \Bbb{C}$, then
$$\lambda_1 = \lambda_2 = \cdots = \lambda_n = 0.$$

 [/INST] In the case of $n=2$ you have
$$
a+b+ab=0\\
a^2+b^2+a^2b^2=0
$$
The second equation transforms to
$$
0=(a+b)^2-2ab+a^2b^2=2(a^2b^2-ab).
$$
The solution $ab=0$ leads to $a=b=0$. The other solution $ab=1$ gives $a+b=1$, so that $a,b$ are the solutions of the quadratic $x^2-x+1=0$ which has non-zero solutions $\frac{1\pm i\sqrt3}2$.
The claim needs some work.
</s>