Let $P:=\{2\ 3\ 5\ 7\ 11\ \ldots\}$ be the set of all primes. Let
$$\Pi_x := \{p\in P : p\le x\}$$
for every positive real number $x$. In note Prime products--a modest beginning I showed a simple proof of inequality
$$\Pi_n\ge n$$
for every natural number $n$. This is an infinitely weaker version of the well known results but it already shows that there are infinitely many primes, and in a simple way (only the original Euclid proof is still simpler while the given inequality seems to prove a little more). Anyway, I'd like to show stronger results along the similar line (while still a far cry from the ambitious classical known theorems).
THEOREM 0 $\forall_{n\ge 10}\ \,\Pi_{\frac n2}\gt n$, where $n$ is understood to be an arbitrary natural number.
PROOF The theorem holds for $n=10$. Now let $n>10$, and--by induction--we assume that the theorem holds for $n-1$, i.e. $\Pi_{\frac{n-1}2}>n-1$. Then
$$\Pi_{\frac n2}\ge n$$
Thus all that we have to prove is that the equality is impossible. Indeed (arguing by a contradiction), assume that:
$$\Pi_{\frac n2} = n$$
Then $n$ is a product of $2$ and of an odd integer (i.e. a product of some odd primes). Thus $n\equiv 2\mod 4$ hence $m:=n-4\equiv 2\mod 4$. It follows that $\frac m2$ is an odd integer. Furthermore, $\frac m2 = \frac {n-4}2 \ge \frac 72 > 1$. Thus there exists an odd prime $p|\frac m2$ hence $p|m$. However $p$ does not divide $4=n-m$ hence $p$ does not divide $n$. This contradicts the equality $n = \Pi_{\frac n2}$ (remember that $p\le \frac m2<\frac n2$). END OF PROOF
We can continue in the same vein:
THEOREM 1 $\forall_{n\ge 20}\ \,\Pi_{\frac n4}\gt n$, where $n$ is understood to be an arbitrary natural number.
PROOF The theorem holds for $n=20$. Let $n>20$, and--by induction--we assume that the theorem holds for $n-1$, i.e. $\Pi_{\frac{n-1}4}>n-1$. Then.
$$\Pi_{\frac n4}\ge n$$
Thus all that we have to prove is that the equality is impossible. Indeed (arguing by a contradiction), assume that:
$$\Pi_{\frac n4} = n$$
Thus $n$ is a product of $2$ and of an odd integer (a product of odd primes), i.e. $n\equiv 2\mod 4$. Let's consider two cases:
Case 1: assume that $m:=n-2$ is not a power of $2$, i.e. there exist a prime odd divisor $p$ of $m$. But $4|m$. It follows that
$$p\le \frac m4 < \frac n4$$
which would mean that $p|n = \Pi_{\frac n4}$. This would mean that $p|4=n-m$, a contradiction. Thus this case is impossible.
Case 2: assume that $n-2 = 2^d$ for certain positive integer $d$. Thus $2^d = n-2\ge 19$, i.e. $d\ge 4$. Define $m:=n-6$. Then
$$m=4\cdot(2^{d-2}-1)$$
is a product of $4$ and of an odd integer $s:=2^{d-2} - 1$, where $s\ge\lceil \frac{n-6}4\rceil \ge\lceil \frac{15}4\rceil =6 $. Thus exponent $d-2$ is at least $3$, i.e. $d\ge 5$, and $s\ge 7$.
Let $p$ be an arbitrary prime divisor of $m$ (i.e. of $s$). Then $p\le\frac m4 < \frac n4$ hence $p|n$. Thus $p|n-m=6$ hence $p=3$. Since odd prime divisor $p$ of $m$ was arbitrary, it follows that $m=4\cdot 3^c$ for certain positive integer $c$. Thus the following equality holds:
$$2^\delta-1 = 3^c$$
where $\delta := d-2$. As can be seen at s.e.d.e., the only solutions $(c\ \delta)$ of the above equation in non-negative integers are $(\delta\ c) = (1\ 0)$ or $(2\ 1)$. However, as noted earlier, $\delta\ge 3$. Thus Case 2 is impossible. Since both cases are impossible, the equality $\Pi_{\frac n4} = n$ never holds for any $n\ge 20$. END OF PROOF
The next challenge starts to show up, namely the question of the initial constants $10\ \,20$ which appear in Theorems 1 2. How do we compute them? Another one is the relation between the prime divisors of $n-d$ and the function $\nu(n)$ which which appears in $\Pi_{\nu(n)}$ -- e.g. we had $\nu(n):=\frac n4$ in Theorem 1. Thus let's consider another simple stage to illustrate the issues more explicitely:
THEOREM 2 $\forall_{n\ge 56}\ \,\Pi_{\frac n8}\gt n$, where $n$ is understood to be an arbitrary natural number.
PROOF The theorem holds for $n=56$. Let $n>56$, and--by induction--we assume that the theorem holds for $n-1$, i.e. $\Pi_{\frac{n-1}8}>n-1$. Then.
$$\Pi_{\frac n8}\ge n$$
Thus all that we have to prove is that the equality is impossible. Indeed (arguing by a contradiction), assume that:
$$\Pi_{\frac n8} = n$$
Thus $n$ is a product of $2$ and of an odd integer (a product of odd primes), i.e. $n\equiv 2\mod 4$. Let's consider four cases:
Case 1: assume that $m:=n-2$ is such that $8|m$ and $m$ is not a power of $2$, i.e. there exists an odd prime divisor of $m$. Then $p\le\frac m8<\frac n8$ hence $p|n$. Thus $p|n-m=2$ which is wrong. Thus case 1 is impossible.
Case 2: assume that $n-2$ is a power of $2$, say $n-2=2^d$ for an integer. Since $n-2\ge 55$ it follows that $d\ge 6$ and $n\ge 66$.
REMARK 0 Cases 1 and 2 cover all possibilities when $8|n-2$.
Let's go back to case 2. Let this time $m:=n-10$. Then $8|m$ and $m$ is not a power of $2$. To be precise,
$$m = 8\cdot(2^{d-3} - 1)$$
where $2^{d-3}-1\ge 7$. Let $p$ be an arbitrary odd prime divisor of $m$. Then $p\le\frac m8 < \frac n8$ hence $p|n$. Thus $p|n-m = 10$. It follows that $p=5$. This means that
$$2^\delta - 1 = 5^c$$
where $\delta = d-3\ge 3$ and $c$ are positive integers. By simple equation 251, the only solution here is $(\delta\ c) = (1\ 0)$. That's a contradiction since $\delta\ge 3\ne 1$. Thus case 2 is impossible.
Case 3: assume that $m:=n-6$ is such that $8|m$ and $m$ is not of the form $2^d\cdot3^c$, i.e. there exists a prime divisor $p>3$ of $m$. Then $p\le\frac m8<\frac n8$ hence $p|n$. Thus $p|n-m=6$, a contradiction. Thus case 3 is impossible.
END OF PROOF