Czech-Polish-Slovak Match

Suppose $x_1,x_2,\dots ,x_n$ satisfy the equations above. Then we have $$\begin{array}{rl}0& =x_1+x_2^2+x_3^3+\cdots +x_n^n-n-(x_1+2x_2+3x_3+\cdots +n{x}_n-\frac{1}{2}n(n+1))\\ & =(x_2^2-2x_2+2-1)+(x_3^3-3x_3+3-1)+\cdots +(x_n^n-n{x}_n+n-1).\end{array}$$ However, the expressions in the brackets are nonnegative. Indeed, for $k\ge 2$ and $x\ge 0$ we have, by the AM-GM inequality, $$x^k+k-1=x^k+1+1+\cdots +1\ge k\cdot \sqrt[k]{x^k}=kx$$ and the equality holds if and only if $x=1$. Therefore we have $x_2=x_3=\cdots =x_n=1$ and, by the first equation, $x_1=1$.