On the sizes of the maximal prime powers divisors of factorials

Let p be any prime, and p⁽νₚ(n!)) the maximal power of p dividing $n!$. It is proved that there exists a positive integer n₀, which depends only on p, such that q⁽νq(n!)) < p⁽νₚ(n!)) for all n ≥ n₀ and all primes $q > p$. For twin primes p and $q = p + 2$ it is proved that the minimal n₀ satisfying q⁽νq(n!)) < p⁽νₚ(n!)) for all n ≥ n₀ is given by n₀ = (p²+p)/2.