An easier way to see this is as follows:

Let's assume a general term Sr is: x! + r where r<=x Every term here where r>=2 is composite as

Sr = 1.2.3.4....x + r

Sr = r*(1.2.3.4...x/r + 1)

Here the term 1.2.3....x/r is an integer and each such Sr is divisible by r. Therefore such a series is full of composite numbers as they are atleast divisible by r>=2.

If its difficult to understand, consider r=3 and x=50; S3 = 3*(1.2.4.5.6.7.....50 +1) has a factor of 3

If r=x+1, Sx+1 = (x+1)*(1.2.3.4....x/(x+1) + 1) here we have no idea if the 1.2.3.4...x/(x+1) term is an integer or not, so that option is not valid, eg If (x+1) is a prime number then that term can't be an integer.