We use Chebyshev's inequality (aka rearrangement). Let $x_1\ge \cdots \ge x_k$ without loss of generality. Note that \[ x_1^{n-1}\ge \cdots \ge x_k^{n-1}\quad\text{and}\quad \frac{x_1}{S-x_1}\ge \frac{x_2}{S-x_2}\ge \cdots \ge \frac{x_k}{S-x_k}, \]where $S=x_1+\cdots+x_k$. So, \[ \sum_i \frac{x_i^n}{S-x_i}\ge \frac1k\sum_i x_i^{n-1} \left(\sum_i \frac{x_i}{S-x_i}\right), \]so it suffices to verify \[ \sum_i \frac{x_i}{S-x_i}\ge \frac{k}{k-1}. \]This is a consequence of A1 from the same shortlist, see my solution here.