![]() Levent's answer stands on its own in answering your question in the negative. ![]() That being the case, $p_n\# 1$ cannot be the prime next larger than $p_n$. As in the case of $p_n\# 1$, each $p_n$ divides $p_n\#$ but none of them divide $1$, so $p_n\#-1$ must either be a prime or have prime factors, in either instance smaller than $p_n\#$. We must now consider the number $p_n\#-1$. Here is how Levent's answer sinks that ship. Your question in effect asks, "Is it ever the case that $p_=p_n\# 1$, that would be the only possible way to answer your question in the positive. Then the smallest number that contains all of those primes is the product of all of those primes, which is called a primorial and is denoted $p_n\#$. A small dilation on the answer of Levent: Let us imagine there is such a number and there are $n$ primes smaller than that number.
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