Elementary proof for $n^2>p>n$ for all $n>1$ [duplicate]

Question

Is there any elementary way of proving that for all natural numbers $n>1$ there exists a prime $p$ such that $n<p<n^2$. And I mean elementary, not using the Prime Number Theorem or Bertrand's Postulate.

Answer 1

Erdos has a nice proof in his paper on the Sylvester Schur theorem from 1934. I will sketch part of it here.

Look at the prime factors of the binomial coefficient $B = \binom{m+n}{n}$. An elementary argument says that the prime $p$ divides $B$ to a power that is less than $m+n$. Therefore $B$ has all its prime factors less than $n$ multiply to a number $C$ that is less than $(m+n)^{\pi(n)}$. Now when $n > 8$ and $m+n \geq n^2$, $B$ gets bigger than the $n$-smooth part of $B$. This means there is a prime bigger than $n$ dividing a number in $(m,m+n]$. Therefore there is a prime in $(n,n^2]$ for $n > 8$. The remaining cases are left to the reader.

For more elementary arguments giving a prime bigger than $n$ dividing $B$ when $m \geq n$, read the 1934 paper. For those interested, I am reworking the arguments to make it more general and still elementary. I invite you to critique a rough draft.

Gerhard "That Simple Enough For You?" Paseman, 2020.07.15.