Let $y_i=\frac{x_i}{A}$. Then $y_0+y_1+y_2+\dots=1$ and hence $y_i \le 1$ so that $y_i^n \le y_i$ and hence \[x_0^n+x_1^n+\dots=A^n\left(y_0^n+y_1^n+\dots\right) \le A^n\left(y_0+y_1+y_2+\dots\right)=A^n\]Equality holds e.g. for $y_0=1, y_1=y_2=\dots=0$ i.e. $x_0=A, x_1=x_2=\dots=0$ so the maximum is indeed $A^n$. On the other hand, by choosing $x_i=\frac{A}{k}$ for $i \le k-1$ and $x_i=0$ we have $x_0^n+x_1^n+\dots=\frac{1}{k^{n-1}} \cdot A^n$. By letting $k \to \infty$, it is clear that arbitrarily small values can be obtained. Finally by continuity every intermediate value clearly can be attained as well. So the set of possible values is the interval $(0,A^n]$. Edit: Of course the second part is only true for $n>1$. For $n=1$, the set of possible values is clearly just $\{A\}$.

I think this sollution is just as the official solution.Nice proof.

Actually there is a little mistake for the maximum because the numbers are required to be strictly positive, so choosing $x_1=x_2=\dots=0$ is not actually allowed. However, it is clear that by making them arbitrarily small, we can get arbitrarily close to $A^n$. So the answer actually is the interval $(0,A^n)$ for $n>1$ and the set $\{A\}$ for $n=1$.