Given an integer $n \geq 3$, determine the maximum value of product of $n$ non-negative real numbers $x_1,x_2, \ldots , x_n$ when subjected to the condition
\begin{align*} \sum_{k=1}^n \frac{x_k}{1+x_k}  =1 \end{align*}
1. **Substitution and Transformation:**
   Let \( \frac{x_i}{x_i+1} = a_i \). This implies that \( a_i \in [0, 1) \) and \( x_i = \frac{a_i}{1 - a_i} \). Given the condition:
   \[
   \sum_{k=1}^n \frac{x_k}{1+x_k} = 1
   \]
   we can rewrite it using \( a_i \):
   \[
   \sum_{k=1}^n a_k = 1
   \]

2. **Product Transformation:**
   We need to maximize the product \( \prod_{i=1}^n x_i \). Substituting \( x_i = \frac{a_i}{1 - a_i} \), we get:
   \[
   \prod_{i=1}^n x_i = \prod_{i=1}^n \frac{a_i}{1 - a_i}
   \]

3. **Using AM-GM Inequality:**
   By the Arithmetic Mean-Geometric Mean (AM-GM) inequality, for non-negative real numbers \( a_1, a_2, \ldots, a_n \):
   \[
   \frac{a_1 + a_2 + \cdots + a_n}{n} \geq \sqrt[n]{a_1 a_2 \cdots a_n}
   \]
   Given \( a_1 + a_2 + \cdots + a_n = 1 \), we have:
   \[
   \frac{1}{n} \geq \sqrt[n]{a_1 a_2 \cdots a_n}
   \]
   Therefore:
   \[
   \left( \frac{1}{n} \right)^n \geq a_1 a_2 \cdots a_n
   \]
   This implies:
   \[
   a_1 a_2 \cdots a_n \leq \left( \frac{1}{n} \right)^n
   \]

4. **Maximizing the Product:**
   To maximize \( \prod_{i=1}^n x_i \), we need to consider the product:
   \[
   \prod_{i=1}^n \frac{a_i}{1 - a_i}
   \]
   Using the inequality \( a_1 a_2 \cdots a_n \leq \left( \frac{1}{n} \right)^n \), we get:
   \[
   \prod_{i=1}^n \frac{a_i}{1 - a_i} \leq \prod_{i=1}^n \frac{\frac{1}{n}}{1 - \frac{1}{n}} = \left( \frac{\frac{1}{n}}{1 - \frac{1}{n}} \right)^n = \left( \frac{\frac{1}{n}}{\frac{n-1}{n}} \right)^n = \left( \frac{1}{n-1} \right)^n
   \]

5. **Conclusion:**
   The maximum value of the product \( \prod_{i=1}^n x_i \) is:
   \[
   \frac{1}{(n-1)^n}
   \]

The final answer is \(\boxed{\frac{1}{(n-1)^n}}\).