Determine all natural numbers $n$ and $k > 1$ such that $k$ divides each of the numbers$\binom{n}{1}$,$\binom{n}{2}$,..........,$\binom{n}{n-1}$
1. **Understanding the Problem:**
   We need to find all natural numbers \( n \) and \( k > 1 \) such that \( k \) divides each of the binomial coefficients \( \binom{n}{1}, \binom{n}{2}, \ldots, \binom{n}{n-1} \).

2. **Using Kummer's Theorem:**
   Kummer's theorem states that the highest power of a prime \( p \) dividing the binomial coefficient \( \binom{n}{m} \) is equal to the number of carries when \( n \) is added to \( m \) in base \( p \).

3. **Analyzing the Binomial Coefficients:**
   For \( k \) to divide each of the binomial coefficients \( \binom{n}{1}, \binom{n}{2}, \ldots, \binom{n}{n-1} \), \( k \) must be a prime power \( p^t \) for some prime \( p \) and positive integer \( t \). This is because if \( k \) were not a prime power, it would have multiple prime factors, and it would be difficult to ensure that all binomial coefficients are divisible by \( k \).

4. **Considering \( n \) as a Power of a Prime:**
   Let \( n = p^t \) for some prime \( p \) and positive integer \( t \). We need to check if \( p \) divides each of the binomial coefficients \( \binom{p^t}{1}, \binom{p^t}{2}, \ldots, \binom{p^t}{p^t-1} \).

5. **Verifying the Divisibility:**
   - For \( \binom{p^t}{m} \) where \( 1 \leq m \leq p^t-1 \), the binomial coefficient \( \binom{p^t}{m} \) is given by:
     \[
     \binom{p^t}{m} = \frac{p^t!}{m!(p^t-m)!}
     \]
   - Since \( p^t \) is a multiple of \( p \), the factorial \( p^t! \) contains \( p \) as a factor at least \( t \) times.
   - The factorials \( m! \) and \( (p^t-m)! \) will not contain \( p \) as a factor \( t \) times unless \( m \) or \( p^t-m \) is a multiple of \( p \), which is not the case for \( 1 \leq m \leq p^t-1 \).

6. **Conclusion:**
   Therefore, \( p \) divides each of the binomial coefficients \( \binom{p^t}{1}, \binom{p^t}{2}, \ldots, \binom{p^t}{p^t-1} \). Hence, the pairs \((n, k) = (p^t, p)\) satisfy the given condition.

\(\blacksquare\)

The final answer is \( \boxed{ (n, k) = (p^t, p) } \) for any prime number \( p \) and positive integer \( t \).