On the sizes of the maximal prime powers divisors of factorials

Let p be any prime, and $p^(\nu_p(n!))$ the maximal power of $p$ dividing $n!$.

It is proved that there exists a positive integer $n_0$, which depends only on $p$, such that $q^(\nu_q(n!)) < p^(\nu_p(n!))$ for all $n \ge n_0$ and all primes $q > p$.

For twin primes $p$ and $q = p + 2$ it is proved that the minimal $n_0$ satisfying $q^(\nu_q(n!)) < p^(\nu_p(n!))$ for all $n \ge n_0$ is given by $n_0 = (p^2+p)/2$.