This one never gets old

I don't understand what you mean by "proof by contradiction not based on whole induction but the argument from a single step". Here is my understanding of the proof, written out in a more rigorous way: 1. The set {2} contains a single prime number. 2. Given a list of n numbers, we can find a number that's relatively prime to those n numbers by multiplying them all and adding 1. Note that this new number may not be prime itself: for example, 2\*3\*5\*7\*11\*13=30,031=59\*509 is not prime, but it is relatively prime to 2, 3, 5, 7, 11, and 13. Also, there is no requirement for the numbers in the list to be prime either: for example, 4\*5\*6+1=121 is relatively prime to 4, 5, and 6 by the same argument (even though 4 and 6 are not prime). 3. We assume the opposite of the premise (i.e., there exists a finite list containing all prime numbers). 4. We apply the construction in part 1 to this finite list, in order to find a number that's relatively prime to all of those numbers. Since this number has no prime factors in common with the prime list, and it's also not 1 (trivially), one of the prime factors of the new number must not be in the list. Thus, we have constructed a new prime which we can add to our list. 5. Repeatedly apply step 4 over and over; whenever you have a finite list of primes, you can always find a new prime that isn't in the list and then add it to the list, so the total number of primes cannot be finite. As I understand it, step 1 is the base case and step 4 is the inductive step. The proof only works due to this inductive structure. First, you prove that there is a list of 1 prime, which implies the existence of a list of 2 primes, which implies the existence of a list of 3 primes, etc. Is there any way to do the proof this way (multiplying everything and adding 1) without induction?