Find all positive integers  $k$ such that for the first $k$ prime numbers $2, 3, \ldots, p_k$ there exist positive integers $a$ and $n>1$, such that $2\cdot 3\cdot\ldots\cdot p_k - 1=a^n$.

[i]V. Senderov[/i]
1. **Trivial Case:**
   - For \( k = 1 \), the first prime number is \( 2 \). We can take \( a = 1 \) and any \( n \) to satisfy the equation \( 2 - 1 = 1^n \). Thus, \( k = 1 \) is a solution.

2. **Assume \( k > 1 \):**
   - We need to find \( a \) and \( n > 1 \) such that:
     \[
     2 \cdot 3 \cdot \ldots \cdot p_k - 1 = a^n
     \]
   - Let \( P_k = 2 \cdot 3 \cdot \ldots \cdot p_k \). Then the equation becomes:
     \[
     P_k - 1 = a^n
     \]

3. **Divisibility by 3:**
   - Notice that \( P_k \) is divisible by \( 3 \) for \( k > 1 \) since \( 3 \) is one of the primes in the product. Therefore, \( P_k \equiv 0 \pmod{3} \).
   - This implies:
     \[
     P_k - 1 \equiv -1 \pmod{3}
     \]
   - If \( n \) is even, then \( a^n \) must be a quadratic residue modulo \( 3 \). However, \(-1\) is not a quadratic residue modulo \( 3 \), which means \( n \) cannot be even. Hence, \( n \) must be odd.

4. **Case 1: There exists an odd prime \( p \mid n \) with \( p \leq p_k \):**
   - Suppose \( p \) is an odd prime divisor of \( n \) and \( p \leq p_k \). Then:
     \[
     a^n + 1 \equiv 0 \pmod{p}
     \]
   - This implies:
     \[
     a^n \equiv -1 \pmod{p}
     \]
   - By Lifting the Exponent Lemma (LTE), we have:
     \[
     \nu_p(a^n + 1) = \nu_p(a + 1) + \nu_p(n) \geq 2
     \]
   - However, \( P_k \) is divisible by \( p \), so \( P_k - 1 \) is not divisible by \( p \), leading to a contradiction.

5. **Case 2: No such prime exists, so \( n > p_k \):**
   - If \( n \) has no prime divisors less than or equal to \( p_k \), then \( n \) must be greater than \( p_k \).
   - Also, \( a \) must be coprime to \( 2, 3, \ldots, p_k \), implying \( a > p_k \).
   - Therefore:
     \[
     a^n + 1 > p_k^{p_k} \geq p_k! \geq 2 \cdot 3 \cdots p_k
     \]
   - This inequality shows that \( a^n + 1 \) is greater than \( P_k \), leading to a contradiction.

Since both cases lead to contradictions, there are no solutions for \( k > 1 \).

The final answer is \( \boxed{ k = 1 } \).