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I've started by letting the $\gcd = d$. With the goal of trying to prove $d$ must be equal to 1. Since $d$ divides both $n!i + 1$ and $n!j + 1$ then there exist integers $k$ and $m$ such that $dk = n!i + 1$ and $dm = n!j + 1$. By rearranging these equations and doing some substitution, I find that $d|i - j $and $d|j$ ai. Am I off base here?

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Now, you were not off base. Note that since $$j-i \equiv 0 \pmod {d}$$ implies that $$n!=\frac{n!}{j-i} \times (j-i)=dm $$ So we have that $$dk=n! \times i+1 \implies dk=dmi+1$$ This implies $d=1$. We are done.

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    How do you make that first conclusion? – 2017-01-20
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    @mmm I just changed the solution to fit what you posted. – 2017-01-20
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    @mmm can you understand the edit? – 2017-01-20
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    Yes that makes sense – 2017-01-20
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    Actually quick question. Can you explain a little bit more about why n! = dm? I'm not quit seeing that. – 2017-01-20
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    @mmm because of the fact that $j-i$ divides $n!$, as it is smaller than $n$, and $j-i$ can be divisible by $d$. – 2017-01-20