Any hints?

There is a unique prime in $(p_1p_2 \dots p_k, p_1p_2 \dots p_{k+1})$ by Bertrand's postulate for each $k = 1, \dots, n-1$.

YaoAOPS wrote: There is a unique prime in $(p_1p_2 \dots p_k, p_1p_2 \dots p_{k+1})$ by Bertrand's postulate for each $k = 1, \dots, n-1$. How do you deal with the $p_n~^{th}$ prime tho

Bluecloud123 wrote: Any hints? Remember the famous proof that there are infinitely many prime numbers YaoAOPS wrote: There is a unique prime in $(p_1p_2 \dots p_k, p_1p_2 \dots p_{k+1})$ by Bertrand's postulate for each $k = 1, \dots, n-1$. It doesn't state that the prime is unique Also, this would imply that there are at least $n$ primes, which is weaker than the problem statement

i guess the construction might be something like this if we can choose 1 prime in the interval [m*p1p2...pk , (m+1)*p1p2....pk] for all m=1,2,.....p(k+1) -1 than we can finish the proof by induction but i dont know how to verify is m*p1p2...pk +1 m=1,2,.....p(k+1) -1 correct or not

math-olympiad-clown wrote: i guess the construction might be something like this if we can choose 1 prime in the interval [m*p1p2...pk , (m+1)*p1p2....pk] for all m=1,2,.....p(k+1) -1 than we can finish the proof by induction but i dont know how to verify is m*p1p2...pk +1 m=1,2,.....p(k+1) -1 correct or not Your idea about $mp_1p_2\ldots p_k +1$ is good, but this number itself can be composite, so you actually need to consider its prime factors.

so hard is there any answer or hint?

Revenge? Or Maybe just not being lazy. First Proof: Note that $p_1, p_2, \dots, p_{n-1}, p_1p_2 \dots p_{n-1} + 1, 2p_1p_2 \dots p_{n-1} + 1, \dots, (p_{n-1})p_1p_2 \dots p_{n-1} + 1$ are all coprime so there must be at least $p_n+n-1$ prime divisors among them. Second Proof: Note that by the Prime Number Theorem, \[ \pi(p_1p_2 \dots p_n) = \pi(\#p_n) > \pi(2^n) > \frac{2^n}{n} > n (\ln n + \ln \ln n) + n + 1 > p_n + n + 1 \]holds for all $n \ge 8$. (The asymptotic number is the stronger $\approx \frac{e^{n}}{\ln n}$ prime numbers I believe.)