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Is there a way to prove this without using Bertrand's postulate?

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    Your question should be clear without the title. After the title has drawn someone's attention to the question by giving a good description, its purpose is done. The title is not the first sentence of your question, so make sure that the question body does not rely on specific information in the title. – Simply Beautiful Art Oct 29 '17 at 21:14
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    Aside from the fact that this claim is not true for $n=1$ ... this very question was asked and answered just a few days ago: https://math.stackexchange.com/questions/2486809/for-all-n-2-prove-there-exists-a-prime-p-such-that-n-p-n/2486815#2486815 – Bram28 Oct 29 '17 at 21:15
  • @Bram28 , regarding the comment you made in the post you linked, would it also be right if I subtracted n! by 2, (n!-2)? Also, is there a proof behind this rule –  Oct 29 '17 at 21:29
  • @ChinoMalaharis No, when you subtract $2$, then the result is divisible by $2$ ... you want to pick $n!-1$, because that one will have to have prime factors greater than $n$. – Bram28 Oct 29 '17 at 21:31
  • @Bram28 So just to clarify, the only numbers that would be divisible by n!-1 would be the n!-1 and 1, making it a prime? –  Oct 29 '17 at 21:36
  • @ChinoMalaharis No, that's not true. For example, $5!-1=119$ is not prime .. but its prime factors $7$ and $17$ are greater than $5$. So, in general, either $n!-1$ is prime itself, or it is not, but if it is not, its prime factors are still greater than $n$ (and of course smaller than $n!$ – Bram28 Oct 29 '17 at 21:47
  • @Bram28 I see... how would you prove that the prime factors of n!-1 are still greater than n in a general term? –  Oct 29 '17 at 22:04
  • @ChinoMalaharis SInce $n!$ is divisble by $2$, the number right before it, $n!-1$ is not. same for $3$, $4$, etc. all the way up and including $n$. So we know that $n!-1$ is definitely not divisible by any numbers from $2$ to $n$ ... so whatever numbers divide $n!-1$, they would have to be grater than $n$, excluding $1$ Of course, but $1$ is not a prime... or in other words: any prime divisor (or prime factor) has to be greater than $n$. – Bram28 Oct 29 '17 at 22:39
  • @Bram28 Ah ok.. and how would you know that the factors of n!-1 are prime? Sorry for asking you so many questions, I just wanna make sure I know this proof well. –  Oct 29 '17 at 22:44
  • Nvm, you don't have to asnwer. I've figured it out. Thank you so much for the help. –  Oct 29 '17 at 22:55
  • @ChinoMalaharis You're welcome! :) – Bram28 Oct 29 '17 at 23:46

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