Let a factorial number be called $f!$. Let the sum of its factors be called $S(f!)$. Let the ratio between the two be “r”, such that $r=\frac{S}{f!}$.
It is conjectured that: $\frac{S(f+1)!}{(f+1)!} = r+0.5$
In other words; going from one factorial to the next increases the ratio between the sum of the factorial's factors and the factorial by $\frac{1}{2}$. I have not yet proven this, and have checked it only until to $f!=120$.
For the purposes of this question the sum will not include the factorial itself but it will include 1. The numbers tested so far are:
$0=\frac{0}{1}$
$.5=\frac{1}{2}$
$1=\frac{1+2+3}{6}$
$1.5=\frac{1+2+3+4+6+8+12}{24}$
$2=\frac{1+2+3+4+5+6+8+10+12+15+20+24+30+40+60}{120}$
You have $$\frac{S(7!)}{7!}=\frac{14304}{5040} = \frac{298}{105}$$ so your conjecture fails.