We need to show that the product of the first k primes, plus 1, is not prime, but has a prime factor larger than any of the first k primes. The first 25 prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97. As per the hint, we will start form k = 2, 3, and so on to get the required result. Note: we have not taken k as 1 since k = 1 is a trivial case. Let k = 2, then 2 \times 3 + 1= 6 + 1 = 7 which is a prime. Let k = 3, then 2 \times 3 \times 5 + 1 = 31 which is a prime. Let k = 4, then 2 \times 3 \times 5 \times 7 + 1 = 211 which is a prime. Let k = 5, then 2 \times 3 \times 5 \times 7 \times 11 + 1 = 2311 which is a prime. Let k = 6, then 2 \times 3 \times 5 \times 7 \times 11 \times 13 + 1 = 30031 = 59 * 509 which is not a prime and is divisible by a prime 59 which is greater than all the prime number 2, 3, 5, 7, 11, and 13. Thus, k = 6 such that the product of the first k primes, plus 1, is not prime, but has a prime factor larger than any of the first k primes. Note: In general, it can be proved that if the product of the k primes, plus 1, is not prime, then has a prime factor larger than any of the taken k primes. Answer: k = 6.