Is this obvious? Starting with the third prime "5",
a prime is less than the product of all preceding primes plus 1 .
A few examples seem convincing that the statement is true, but I have not been able to demonstrate it.
Yes it's true. Google "Bertrand's Postulate", so that if we have the prime p and p' is the prime before it, then p < 2p'...
And so, in fact, your statement could very seriously be strengthened: every prime >= 3 is less than the prime before it times the first prime.
No, not obvious, but pretty straigthforward for whoever knows BP.
Tonio
No need for Bertrand's postulate.
Simply note that $\displaystyle p_1\cdot ...\cdot p_n-1$ is coprime to $\displaystyle p_1,...,p_n$, therefore it must be divisible by another prime (since it's >1) which is not among the first n, since the sequence of primes is increasing (*) we conclude that $\displaystyle p_{n+1}\leq p_1\cdot ...\cdot p_n-1<p_1\cdot ...\cdot p_n+1$
(*) $\displaystyle p_{n+k}|(p_1...p_n-1)$ for some $\displaystyle k\geq 1$ hence $\displaystyle p_{n+1} \leq p_{n+k}\leq p_1...p_n-1$
Thanks, Tonio and PaulRS! I shall look up Bertrand's Postulate, Tonio.
PaulRS: Yes, now that I have your reply, I see the simplicity I was not able to conger. G.E. Andrews, in his "Number Theory" states the inequality in a hint towards the solution of an exercise, but I could not demonstrate the hint - let alone the entire exercise!
Thanks to you, both. Now, I can struggle on.
Nivla.