Using generalized factorial(or multifactorial or k-torial) we obtain:
1:
100=11!!!!!!+11!!!!!!!+1
2:
100=22!!!!!!!!!!!!!!!!!!+(2+2)!/2
3:
100=((3!)!!)!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!+3!-(3!+3!)/(3!)
4:
100=(4!)!!!!!!!!!!!!!!!!!!!!+4+(4-4)/4
5:
100=(5!!)!!!!!!!!!+5+5+5-5
6:
100=(6!!)!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!+6-(6+6)6
7:
100=(7!!!)!!!!!!!!!!!!!!!!!!!!!!!!!+7!!!!!+(7+7)/7
8:
100=(8!!!!+8!!!!!!+(8+8)/8)!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
9:
100=((9+9/9)!!!!!)!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!+9-9
So we got 100 using all the i's from 1 to 9.
Edit:
I found a formula that works for a any integer greater than 9:
$$100=\left(\left(\left(\frac{\left(\frac{(n+n){!}_{(n)}}{n+n}\right){\large!}_{(n-5)}}{n}\right){\huge!}_{(3)}\right){\huge!}_{(5)}\right){\huge!}_{(48)}$$ where $a!_{(b)}=a\overbrace{!!!\cdots!!!}^{b\mbox{ times}}$
We can write a formula for every number without using $.i$