[Ask]prove that for all integers 'n', if n is bigger than 2. then there is a prime number 'p' between n and n! I think using contradiction will be useful.
Suppose $p$ exists which divide $n!-1$
and suppose that
 $p \leq n$ leads to a contradiction.
But I have no idea how to show contradiction
I believe someone can help me.(I'm not English-speaker also)
 A: This is essentially getting at a variant of Euclid's proof that there are infinitely many primes.
Given $n > 2$, you just multiply all the numbers from 2 to $n$, be they prime or not. We notate this product $n!$. Whatever primes there are between 2 and $n$, we know that $n!$ is divisible by all of them.
But $n! - 1$ is not divisible by any of those primes: $n! - 2$ is divisible by 2, $n! - 3$ is divisible by 3, etc., up to the largest prime equal to or less than $n$.
In your book there should be a theorem and proof that every integer greater than 1 is either prime or the product of primes. Then $n! - 1$ must be either prime or the product of primes.
So if $n! - 1$ is prime, then we have found a prime between $n$ and $n!$. But if $n! - 1$ is composite, we need to go one step further. Let's say $n! - 1 = pq$, where $p$ is a prime, $q$ may or may not be a prime, and $p < q$.
Let's also say that $p < n$. But we already saw that $n!$ is divisible by all the positive primes up to $n$, and so $n! - 1$ is divisible by none of them, so we have contradicted ourselves by asserting that $p < n$, which is now shown to be wrong.
Therefore $n < p < q < n! - 1$, again showing the existence of at least one prime between $n$ and $n!$.

There is a much stronger result: there is always a prime between $n$ and $2n$ for $n > 1$, and clearly $2n < n!$ for $n > 3$. This is Bertrand's postulate, but even the "elementary" proofs are too involved to repeat here.
