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 $n_0=(p^2+p)\mathrm{/}2$.