In case 1 when I said, "it is easy show that every prime factor...", I have to admit that I was thinking specifically about non-repeated prime factors at the time, in which case the result really is trivial enough to be done "by inspection".
For the case where (2n+1) has repeated prime factors like q^m then you can use an arguement along the lines of :
Assume (2n+1) has a prime factor of the form q^m, where q is prime >= 3 and m is integer >= 2.
BTW. Note that (2n+1) can't have 2 as a factor so we only need to look a 3 and greater. Also, since I'm looking specifically at the case of repeated factors I only need to consider m greater or equal to 2.
Since q^m is a factor then, (2n+1) >= q^m
(2n-1) >= q^m - 2
But also it is easy to show that q^m - 2 > m*q for q>=3 and m>=2
Loosly what this means is that if (2n+1) has a repeated prime factor q^m then (2n-1)! has enough factors of the form q, 2q, 3q etc to "cover" it.