1. Applying Chebyshev's Inequality:

Chebyshev's Inequality states that for sequences $\{a_i\}$ and $\{b_i\}$ that are both increasing or both decreasing, and real numbers $\mu_i$ $(1 \leq i \leq n)$, we have:
$$\sum_{i=1}^n \mu_i a_i \sum_{i=1}^n \mu_i b_i \leq \sum_{i=1}^n a_i b_i \sum_{i=1}^n \mu_i$$

2. Setting up the sequences:

Let $a_i = x_i^k$ and $b_i = x_i^{n-1}$. Since $x_i > 0$, both $a_i$ and $b_i$ are increasing sequences if $x_i$ are increasing. We also set $\mu_i = 1$ for all $i$.

3. Applying Chebyshev's Inequality:

Using Chebyshev's Inequality with the sequences and $\mu_i$ as defined:
$$\left( \sum_{i=1}^k 1 \cdot x_i^k \right) \left( \sum_{i=1}^k 1 \cdot x_i^{n-1} \right) \leq \left( \sum_{i=1}^k x_i^k x_i^{n-1} \right) \left( \sum_{i=1}^k 1 \right)$$
Simplifying, we get:
$$\left( \sum_{i=1}^k x_i^k \right) \left( \sum_{i=1}^k x_i^{n-1} \right) \leq \left( \sum_{i=1}^k x_i^{n+k-1} \right) k$$

4. **Dividing by $k$:**

Dividing both sides by $k$:
$$\frac{1}{k} \left( \sum_{i=1}^k x_i^k \right) \left( \sum_{i=1}^k x_i^{n-1} \right) \leq \sum_{i=1}^k x_i^{n+k-1}$$

5. Using AM-GM Inequality:

By the Arithmetic Mean-Geometric Mean (AM-GM) Inequality, we have:
$$\prod_{i=1}^k x_i \leq \left( \frac{1}{k} \sum_{i=1}^k x_i \right)^k$$
Applying this to our context:
$$\prod_{i=1}^k x_i \leq \frac{1}{k} \sum_{i=1}^k x_i^k$$

6. Combining the results:

Combining the results from Chebyshev's Inequality and AM-GM Inequality, we get:
$$\prod_{i=1}^k x_i \cdot \sum_{i=1}^k x_i^{n-1} \leq \sum_{i=1}^k x_i^{n+k-1}$$

This completes the proof.

$\blacksquare$