There is a prime between $n$ and $n^2$, without Bertrand Consider the following statement:

For any integer $n>1$ there is a prime number strictly between $n$ and $n^2$.

This problem was given as an (extra) qualification problem for certain workshops (which I unfortunately couldn't attend). There was a requirement to not use Bertrand's postulate (with which the problem is nearly trivial), and I was told that there does exist a moderately short proof of this statement not using Bertrand. This is my question:

How can one prove the above statement without Bertrand postulate or any strong theorems?

Although I can only accept one answer, I would love to see any argument you can come up with.
I would also want to exclude arguments using a proof of Bertrand's postulate, unless it can be significantly simplified to prove weaker statement.
Thank you in advance.

I've come up with a simple proof based on the prime-counting function $\pi(x)$, which I'm pretty sure doesn't depend on Bertrand's Postulate.
First, I will prove a lemma that for every prime $n$, there is another prime $p$ with $n < p < n^2$. I will use this result later to show the general result (i.e. for composite $p$ as well).
Based on some inequalities of $\pi(x)$ and the value of a related constant we have
$$\frac{n}{\log n} < \pi(n) < C\frac{n}{\log n}\text{, where }C=\pi(30)\frac{\log 113}{113} \approx 1.255$$
for all $n \geq 17$. We want to prove that for all prime $n$ we have
$$\pi(n^2) > \pi(n)$$
or
$$\pi(n^2) - \pi(n) > 0,$$
so we look at the upper bound of $\pi(n)$ and the lower bound of $\pi(n^2)$ to find a minimum difference.
We have
$$\begin{align}
\pi(n^2) - \pi(n) &> \frac{n^2}{\log n^2} - C\frac{n}{\log n} \\
&= \left(\frac{n}{2} - C\right)\frac{n}{\log n}
\end{align}
$$
so the conclusion is satisfied whenever
$$\left(\frac{n}{2} - C\right)\frac{n}{\log n} > 0.$$
Since for all prime $n$ we have
$$\frac{n}{\log n} > 0,$$
we find that $\pi(n^2) - \pi(n) > 0$ whenever $\frac{n}{2} - C > 0$, which gives us
$$n > 2C \approx 2.51$$
so $$n \geq 3$$
which is true for every prime $n \geq 17$.
For lesser primes we can prove things manually:
$$2 < 3 < 2^2$$
$$3 < 5 < 3^2$$
$$5 < 7 < 5^2$$
$$7 < 11 < 7^2$$
$$11 < 13 < 11^2$$
$$13 < 17 < 13^2$$
This concludes the lemma.
Then, we must prove that for every composite $n$ there is a prime $p$ with $n < p < n^2$. Take the largest prime less than $n$ and call that $p_m$. Then by applying the lemma we find that there is some prime $p$ such that
$$p_m < p < p_m^2.$$
Because $p > p_m$, $p_m$ is the largest prime less than $n$, and $n$ is composite, we find that 
$$p < n.$$
Also, we have
$$p < p_m^2 < n^2,$$
so we reach the second conclusion: if $n$ is composite then there's a prime $p$ such that
$$n < p < n^2.$$
Now we've proved the original statement for both prime and composite $n$, and the proof is complete.
