Daniel Roca González: Yes. Why are you not specific here. Just pick one, say 2. Why should there be 3 different prime factors? IMO you are not giving a complete argument here. If you spell out the details to make it rigorous by modern standards, you will likely need more then half a page. Euclids proof, on the other hand is super clean and concise. Euclid's Theorem is one of the fundamental results in number theory, which asserts that there are infinitely many prime numbers. The first proof of Euclid's Theorem was in the 9th book of Euclid's geometric treatise ( Elements) in 300 B.C. .  Figure: The frontispiece of Sir Henry Billingsley's first English version of Euclid's Elements, 1570 While Euclid's proof is usually reworded like this, it is not actually necessary to use contradiction in this way. Instead of assuming P is the finite set of all prime numbers, you can take $P$ to be any finite set of primes, and follow in the same way. The proof shows that $Q$ has a prime factor which is not in $P$, which already proves the premise since $P$ was arbitrary. You can use this iteratively(just put the new prime in the set $P$ and apply the proof again to get a new prime) to generate an infinite set of primes, just like Saidak's proof, and it's just as simple. Besides being a concise constructive proof, Saidak's relies solely on the property that consecutive integers are coprime and that's what makes this proof unique. Two numbers are coprime if their highest common factor (or greatest common divisor) is 1. Thx @Erik for spelling it out. I think this argument deserves a little more elaboration than just "Similarly" in a proof of an elementary result, that should be digestible by beginners. Saidak's new proof seems awfully similar to his restatement of Euclid's proof. In both cases the key is that any finite set of primes $\{p_1, p_2, ..., p_n\}$ can be extended by appending the prime factors of $p_1p_2(...)p_n+1$, which don't belong to ${p_1, ..., p_n}$. $Q = p_1.p_2...p_k + 1= P+1 $ is either prime or not. 1. If Q is prime then by contradiction there's one more prime in the set P 2. If Q is not prime, there is a $p_i$ in the set P that divides Q. By definition, since $p_i$ is in the list it will also divide P. If $p_i$ divides Q and P it will also divide its difference $Q-P=1$, which is a contradiction since no prime number divides 1. This means that $p_i$ is a prime number and can't be in the list! As can be seen, Euclid proves the result by contradiction, assuming that the initial set P contains all the prime numbers. Saidak's proof unlike Euclid's avoids the "reductio ad absurdum" method. And even if there were three, why can't two of them coincede? E.g. $2 \cdot 2 \cdot 3$ ? You are not giving a complete argument here. Coprime means that two numbers share no common prime factors. $n$ and $n+1$ share no prime factors, so $n(n+1)$ must have at least 2 prime factors. Consequently, the number after $n(n+1)$, $n(n+1)+1$ shares no common factors with $n(n+1)$, so it must share no common factors with either $n$ or $n+1$ since you can't get new prime factors by multiplying together $n$ and $n+1$ This is another incomplete argument. Be precise: "which process" and why are the generated numbers different from each other? In this paper Filip Saidak, Professor at The University of North Carolina at Greensboro, comes up with a proof that is conceptually even simpler than Euclid's proof. Euclid's Theorem has been proved by several other famous mathematicians including Euler and Paul Erdos. Euler's and Erdo's proofs relied on the fundamental theorem of arithmetic: that every integer has a unique prime factorization. To see their proofs [click here](https://en.wikipedia.org/wiki/Euclid%27s_theorem#Proof_using_factorials) @Heinrich The implicit claim is that $N_i$ contains at least $i$ different prime factors. $N_i + 1$ is coprime with $N_i$, so it has $\geq 1$ prime factor different from the $\geq i$ different prime factors in $N_i$. Thus, $N_{i+1} := N_i (N_i + 1)$ has $\geq i+1$ different prime factors. Together with the base case that $N_1 := n > 1$ has at least one prime factor, this proves the claim via induction.