Let p be any prime, and $p^(ν_p(n!))$ 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 $q^(ν_q(n!)) < p^(ν_p(n!))$ 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 $q^(ν_q(n!)) < p^(ν_p(n!))$ for all $n \ge n_0$ is given by $n_0 = (p^2+p)/2$.