Actual Question:
Show that if the positive integers $n$ and $n+k$, where $n>k$ and $k$ is an even positive integer, are both prime, then $(k!)^2((n-1)!+1)+n(k!-1)(k-1)!\equiv0$ mod $n(n+k)$
I asked this question yesterday (albeit poorly worded so I deleted and decided to re-ask the question with clarity). So far I have: $$A=(k!)^2[(n-1)!+1]+(n+k)(k!-1)(k-1)!$$ For the original statement to be true, I'd have to show: $$n(n+k)|A$$ Which means more reducing/ canceling/simplifying. Neither of which am I good at. My question is, what else can I do to be able to shrink down $A$ so I can show $n(n+k)|A$ to prove that the original statement holds true?