Comparing Two Gigantic Numbers

Notice that $887$ is prime. Wilson's theorem states that $(p-1)! \equiv -1 \pmod p$ where $p$ is a prime number. Therefore $886!+1$ is divisible by $887$ and so is $887!$

This is indeed the greatest common factor which is possible as $\gcd(887!,886!+1)=\gcd(887!,887n)=887×\gcd(886!,n)=887$ as for some integer $n$ ,

$887n$ divides $886!+1$ , an odd number which implies $n$ is odd and the result.

First, we got: $k|886!(\forall k \in \overline{1;886})$ So $886!+1\equiv 1 \pmod {k}$ And apply Wilson's Theorem, we have: $886!+1\equiv (-1+1)\equiv 0 \pmod {887}(because~887~is~prime)$ So the only common factor, which is also the answer, is $\boxed{887}$