Quote:
Originally Posted by bur
A sum will never have a prime factor that isn't a prime factor of all addends. Since for all primes <= pn there is a addend that doesn't contain it, the sum, i.e. the numerator, will not be divisible by any of p1 ... pn.

This is not
quite correct. The important point is that p
_{i} divides _all_ terms except for _one_. If more than one term was not divisible, you can't say anything about divisibility of the whole sum. But in this case, we know that everything else is divisible by p
_{i}, and therefore the whole is _not_ divisible.
Rest of the argument is fine, I think. You can't simplify that fraction any further (since none of the prime factors of the denominator appears in the numerator, by the above argument).