Can we find $x_{1}, x_{2}, ..., x_{n}$? Consider this.
$$x_{1}+x_{2}+x_{3}+....+x_{n}=a_{1}$$
$$x_{1}^2+x_{2}^2+x_{3}^2+....+x_{n}^2=a_{2}$$
$$x_{1}^4+x_{2}^4+x_{3}^4+....+x_{n}^4=a_{3}$$
$$x_{1}^8+x_{2}^8+x_{3}^8+....+x_{n}^8=a_{4}$$
$$.............................$$
$$x_{1}^{2^{n-1}}+x_{2}^{2^{n-1}}+x_{3}^{2^{n-1}}+....+x_{n}^{2^{n-1}}=a_{n}$$
Can we find $x_{1}, x_{2}, ..., x_{n}$ by knowing $a_{1}, a_{2}, ..., a_{n}$?
 A: For $n \ge 3$ a solution (if it exists) need not be unique. As an example,
with $\eta = e^{2\pi i /3}$, both $(0, 0, 0)$ and $(1, \eta, \eta^2)$
satisfy
$$
  x_1 + x_2 + x_3 = 0 \\
  x_1^2 + x_2^2 + x_3^2 = 0 \\
  x_1^4 + x_2^4 + x_3^4 = 0 
$$
Similarly, both $(0, \ldots, 0)$ and $(1, \eta, \eta^2, 0, \ldots, 0)$
solve the system for any $n > 3$ with $a_1 = \ldots = a_n = 0$.
For $n = 2$ the solution can be computed explicitly by
solving a quadratic equation, and for $n=1$ it is trivially true.

Also a solution need not exist. Here is an example for $n=3$:
You have
$$
 a_1 = p_1, a_2 = p_2, a_3 = p_4
$$
where $p_j$ are the "power sums"
$$
 p_k = x_1^k + x_2^k + x_3^k \, .
$$
These are related to the "elementary symmetric polynomials"
$$
\begin{aligned}
 e_1 &= x_1 + x_2 + x_3 \\
 e_2 &= x_1 x_2 + x_2 x_3 + x_3 x_1 \\
 e_3 &= x_1 x_2 x_3 \\
 e_k &= 0 \text{ for } k \ge 4
\end{aligned}
$$
via "Newton's identities",
which for $n= 3$ are
$$
\begin{aligned}
 p_1 &= e_1 \\
 p_2 &= e_1 p_1 - 2 e_2 \\
 p_3 &= e_1 p_2 - e_2 p_1 + 3e_3\\
 p_4 &= e_1 p_3 - e_2 p_2 + e_3 p_1
\end{aligned}
$$
Now let us assume that $a_1 = p_1 = 0$. Then it follows that
$$
\begin{aligned}
  p_2 &= - 2 e_2 \\
 p_4 &=  - e_2 p_2 
\end{aligned}
$$
and therefore $p_4 = \frac 12 p_2^2$.
In other words, if $a_1 = 0$ and $a_3 \ne \frac 12 a_2^2$ then 
there is no solution.
A: If you mean the diophantine-equation tag and the solutions are supposed to be integers, searching will be easy because high powers are spaced far apart.  Because of the symmetry you can insist that $x_1 \ge x_2 \ge \dots \ge x_n$.  You can focus just on the last equation $x_{1}^{2^{n-1}}+x_{2}^{2^{n-1}}+x_{3}^{2^{n-1}}+....+x_{n}^{2^{n-1}}=a_{n}$.  We have $x_{1}^{2^{n-1}} \le a_n \le nx_{1}^{2^{n-1}}$ or $(\frac{a_n}n)^{2^{1 \le -n}}  x_1 \le a_n^{2^{1-n}}$  Since $n^{2^{1-n}}$ is just a bit greater than $1$, this is a tight bound.  
For example, let $n=5$ and the final equation be $x_{1}^{16}+x_{2}^{16}+x_{3}^{16}+x_{4}^{16}+x_{5}^{16}=46512832447930819$  We get $9.955 \le a_1 \lt 11.008$, so $a_1$ must be $10$ or $11$.  If we take $a_1=10$, then $9.94 \lt a_2$, so $a_2=10$ as well.  It turns out we get driven to $10,10,10,10,10$, which is too large, or $10,10,10,10,9$, which is too small.  If we take $a_1=11$ things work better (because I made them).  We get $7.66 \lt a_2 \lt 8.35$, so $a_2=8$, then $7.47 \lt a_3 \lt 8.002,$ so $a_3=8$ and we find $a_4=5, a_5=3$ works.
