Past exam of the options II course of the University of Oxford / 2005 / C / 2 / b / iii / Solution

If a prime divisor $p$ of $(n!{)}^2+1$ satisfied $p\le n$, it would divide $n!$ and therefore divide the difference $(n!{)}^2+1-(n!{)}^2=1$, impossible. Hence $p>n$.

Every integer $(n!{)}^2+1>1$ has a prime divisor, and (ii) makes each of its prime divisors congruent to 1 modulo 4. Taking arbitrarily large $n$ produces such primes beyond every bound. ThereforeEquivalently, if their list were finite, choosing $n$ at least its largest member would produce a new member larger than that supposed maximum.