It is possible to shorten Bertrand's Postulate's proof so it proves only the above. We can throw away the usually-proven upper bound on the primoral. Explicitly, following Wikipedia's "Proof of Bertrand's postulate":

Lemma 1: $$\frac{4^{\lfloor n^2/2 \rfloor}}{2\lfloor n^2/2 \rfloor+1} < \binom{n^2}{\lfloor n^2/2 \rfloor}$$ For a fixed prime $p$, define $R(p,n)$ to be the highest natural number $x$, such that $p^x$ divides $\binom{n}{\lfloor n/2 \rfloor}$.

Lemma 2: $$p^{R(p,n)} \le n+1$$

If there are no primes between $n$ and $n^2$, then: $$\binom{n^2}{\lfloor n^2/2 \rfloor } = \prod_{p\le n} p^{R(p,n^2)} < (n^2+1)^n$$

This violates lemma 1 as soon as $n \ge 7$.

(* the floors where put in a bit hastily)