Factorial Way

Using the notation $\gcd(x,y)$ to denote the greatest common divisor (a.k.a. highest common factor) of $x$ and $y,$ we can write

$\begin{aligned} H &= \gcd(n!+1, (n+1)!) \qquad \qquad \\ &= \gcd(n!+1, n!(n+1)) \\ &= \gcd(n!+1,n+1) & \text{Since } \gcd(n!+1, n!) = 1 \\ &= \begin{cases}n+1 & \text{ if } (n+1) \text{ is a prime} \\ 1 & \text{ otherwise}\end{cases} & \text{By Wilson's theorem} \end{aligned}$

The problem can then be rephrased to "What is the largest prime $p = n+1 \leq 100$ ?" and therefore the answer is $\boxed{97}$