A question about prime factorization of $n!$

Question

Prove that for any integer $K$, There exists a natural number $N$ so that in the prime factorization of $N!$ we can find at least $K$ prime numbers which their powers are exactly $1$.

Answer 1

The primes in $(N/2,N]$ appear with exponent 1, so you need to show that there are arbitrarily many primes in such an interval. You could use Ramanujan's proof of Bertrand's postulate, for example.

If you have access to sledgehammers like the Prime Number Theorem that would suffice as well.

Answer 2

Hint Prove that there exists some $n$ so that $\pi(2n)-\pi(n)=K$, where $\pi$ is the prime counting function. Then you can show that $(2n)!$ has the desired result.

To prove this hint, try combining the Prime Number Theorem with the fact that $f(n)=\pi(2n)-\pi(n)$ satisfies

$$f(n+1)-f(n) \in \{ 0, 1, -1 \} \,;\, \forall n\ geq 3 \,.$$