Prime number between $n$ and $n!+1$ I am trying to prove that ($\forall \ n\in\mathbb{N}$) there exists a prime number $q$ such that $n < q \le 1 + n!$
I have made a graph with $n=0$ through $n=10$ and found solutions to all of them looking for a pattern and I see that $n!$ gets enormous fast and it becomes quite obvious that there is a prime number in between them.
I have considered trying to prove by contradiction that $q$ does not exist on that interval, but I don't know where to go from that statement.  Could anybody help me figure it out?  I have been staring at it for hours and I can't figure out where to go.
Thank you.
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 A: All the primes dividing $n!$ give remainder $1$ when they divide $n!+1$. Those include all primes from $1$ to $n$. So either $n!+1$ is itself a prime, or it is divisible by a prime $>n$ and of course $\le n!+1$.
A: Hint: $n!+1$ has some prime factor $p$. If $p \leq n$ then $p\mid n!$.
A: For $n=1$ and $n=2$ the condition holds.
So to prove the statement assume $n>2$. Now for every integer $x$ such that $1<x<(n+1),$ we have $x|n!$ and $x\not|(n!-1).$ 
$\therefore$ either $(n!-1)$ is a prime, or $\exists$ a prime $p\ge (n+1) $ such that $p|(n!-1)$.
So in any case, $\exists$ a prime $p$ such that $(n+1)\le p\le (n!-1)$.
$\therefore$  $\exists$ a prime number $p$ such that $n<p≤1+n!$ $\hspace{.2cm}$$,\forall n\in \mathbb{N}.$
A: $n!+1$ isn't divisible by the integers from $2$ to $n$. Then all its prime factors exceed $n$, and there is at least one.
A: Case 1) if n=1,2,3 then $q=2,3,7$
Case 2) for other n, $2n< n! $ Now we know $\exists q$ prime s.t $n<q<2n<1+n!$
Hence we are done. 
A: HINT:
Use the Bertrand's postulate.
Since $n!\ge 2n$ for all $n\ge 3$ we have the result.
