We have for otherwise , that term would not be a prime .
Also because is not a prime .
Let's consider the lemma ,
is divisible by provided that is prime to
Take modulo and since is prime to it , there is an inverse of it , let
, it is deduced by using the fact that is an integer .
It should be divisible by if is prime to but we have each term is prime , we have the product contains prime factors that all are greater than , which is prime to . Therefore , is prime to is false for and we have that contains all prime .
Consider , ,
contains these two prime numbers but not also . It is not true that every prime , must divide a , it should be
I find that i can simplify the solution i gave ... Maybe the first thing is to modify the lemma i introduced ...
is divisible by where provided that is prime to .
The proof is quite similar to that of the original lemma .
Then , we just need to consider the product :
Let be the prime not exceeding ie .
It is divisible by if is prime to . However , each of the terms is prime so it doesn't contain the prime , a contradiction makes us to say ' is prime to ' is false and .
Note: We can also say for if . In the counterexample I gave , it is coincident that . Not being in this case , the statement is true .