Show that for all integer n 0, there exists a prime p such that n

Question

Show that for all integer n >0, there exists a prime p such that n < p =< n!+1
Please, help

irishboy123 · Answer

this is an induction proof, right?

loser66 · Answer

I am not sure about that, but if it helps, why not, right?

loser66 · Answer

I can prove the LHS and the middle like
Let p is an arbitrary prime, then (p, 1) =1. By Bezouth theorem, in the sequence $\{pn+1\}_{n=1}^\infty $ there are infinite many prime, hence we always have a prime under that form pn +1
Surely, with integer n>0, n < pn +1. 
But not sure about the far right.

irishboy123 · Answer

really sorry loser, i am crap at this pure  math stuff;  but i did find Bertrand's Postulate:

 for every $n > 1$, here is always at least one prime p such that:

$$n < p < 2n$$

interesting stuff:-)

loser66 · Answer

Hey, is it not that it is done? since 2n < n! = 1*2*3*....*n <n! +1
Wow!! unfortunately, if I use it, I have to prove Bertrand's postulate, right?

loser66 · Answer

And the book said Bertrand could not produce a proof $\implies$ I cannot either. :)

loser66 · Answer

@dan815

loser66 · Answer

@amistre64

loser66 · Answer

@IrishBoy123  , please, tell me |dw:1443050900999:dw|