The bigger factorial mess!

We can generalize the answer with this: Let $n$ be the largest number of a fraction like this

Where the numbers are consecutive. We can factor the terms of each product considering that $n!=n(n-1)!$ and $(n-k)!=(n-k)(n-k-1)!$ with $k\in Z$ and $0<k<n$ , with this and simplifying the fraction that equals to

Given that the fraction equals $n+1$ we can solve this problem easily. In this case, $n=100$ and the fraction have consecutive numbers (without the factorial), so the answer must be $n+1=100+1=101$

Answer is 100 + 1 = 101

Check my solution on the easy version, we can conclude that the asnwer is going to be $100+1=101$