For a given integer $k \geq 1$, find all $k$-tuples of positive integers $(n_1,n_2,...,n_k)$ with $\text{GCD}(n_1,n_2,...,n_k) = 1$ and $n_2|(n_1+1)^{n_1}-1$, $n_3|(n_2+1)^{n_2}-1$, ... , $n_1|(n_k+1)^{n_k}-1$.
1. **Initial Assumption and Simplification**:
   - We are given that $\text{GCD}(n_1, n_2, \ldots, n_k) = 1$.
   - We need to find $k$-tuples $(n_1, n_2, \ldots, n_k)$ such that:
     \[
     n_2 \mid (n_1 + 1)^{n_1} - 1, \quad n_3 \mid (n_2 + 1)^{n_2} - 1, \quad \ldots, \quad n_1 \mid (n_k + 1)^{n_k} - 1
     \]

2. **Case Analysis for $n_i = 1$**:
   - If any $n_i = 1$, then:
     \[
     n_{i+1} \mid (1 + 1)^1 - 1 = 1
     \]
     This implies $n_{i+1} = 1$. By induction, if one of the $n_i$ is 1, all $n_i$ must be 1. Thus, $(1, 1, \ldots, 1)$ is a solution.

3. **Assume All $n_i > 1$**:
   - Let $p_i$ be the smallest prime divisor of $n_i$.
   - We need to show that this leads to a contradiction.

4. **Order Argument**:
   - Consider the condition $n_{i+1} \mid (n_i + 1)^{n_i} - 1$.
   - By properties of orders in modular arithmetic, the order of $(n_i + 1)$ modulo $n_{i+1}$ must divide $n_i$.
   - If $p_{i+1}$ is the smallest prime divisor of $n_{i+1}$, then $p_{i+1} \mid (n_i + 1)^{n_i} - 1$.

5. **Prime Divisor Analysis**:
   - If $\text{gcd}(p_{i+1} - 1, n_i) > 1$, there exists a prime $q$ such that $q \mid n_i$ and $q \mid p_{i+1} - 1$.
   - This implies $p_i \leq q \leq p_{i+1} - 1$, hence $p_{i+1} > p_i$.

6. **Cyclic Argument**:
   - If $p_{i+1} \geq p_i$ for all $i$, then by cycling through all indices, we get $p_1 \leq p_2 \leq \ldots \leq p_k \leq p_1$.
   - This implies $p_1 = p_2 = \ldots = p_k$, meaning all $n_i$ share a common prime factor, contradicting $\text{GCD}(n_1, n_2, \ldots, n_k) = 1$.

7. **Conclusion**:
   - The only possible $k$-tuple that satisfies all conditions is $(1, 1, \ldots, 1)$.

The final answer is $\boxed{(1, 1, \ldots, 1)}$