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Assuming that $p_{n}$ is the $n^{ th}$ prime and $p_{n}\text{#}$ is the $n^{ th}$ primorial, what is a proof in elementary number theory that, for all $n \ge 2$, $p_{n+1} \le p_{n}\text{#}$ ?

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Bertrand’s postulate isn’t really an elementary approach. Consider instead the decomposition into primes of the number $p_n\#-1$ (which is greater than $1$ when $n\ge 2$). This number is coprime with any $p_i,$ where $i\le n.$ So all primes in this decomposition (there is at least one) are greater than $p_n$. And less than $p_n\#$.

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By Bertrand's postulate, there exists a prime between $p_n$ and $2p_n$ such that:

$p_{n+1}<2p_n$

And now we need to prove that

$2p_n \le p_n $# wich is clearly true when $n\ge 2$

And the result follows immediately

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