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