Proof that ∀n∈ℕ(n>2⟹∃p∈ℙ(n<p<n!))

Prove that, for every natural number n greater than 2, there is a prime number between n and n!.

Proof: For every natural number $n>2$, there is a prime number between $n$ and $n!$.

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Step 1: Understand the problem.

We are asked to show that for any integer $n>2$, there exists at least one prime $p$ such that:

$$n<p<n!$$

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Step 2: Use a classical approach based on divisor arguments.

Consider the number:

$$N=n!-1$$

Let's analyze the prime factors of $N$.

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Step 3: Key observation.

• For any integer $k$ with $2\le k\le n$, $k$ divides $n!$ (by the definition of factorial).
• But $k$ does not divide $n!-1$ because:

$$n!\equiv 0(\mathrm{mod}k)⟹n!-1\equiv -1\not\equiv 0(\mathrm{mod}k)$$

This means none of the integers $2,3,\dots ,n$ divides $N=n!-1$.

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Step 4: Consequence.

• Any prime divisor of $N$ must be greater than $n$.

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Step 5: Conclusion.

• $N=n!-1$ is either prime itself (in which case it's a prime greater than $n$) or composite.
• If it's composite, it must have a prime factor greater than $n$, since none of the primes $\le n$ divide it.

Thus, there exists a prime $p>n$ such that $p$ divides $n!-1$.

Since $n!-1<n!$, this prime $p$ satisfies:

$$n<p<n!$$

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✅ Therefore, for every integer $n>2$, there is a prime number between $n$ and $n!$.

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This proof uses an idea similar to Euclid's proof of the infinitude of primes but applied to the interval $(n,n!)$.