Using [generalized factorial(or multifactorial or k-torial)](https://en.wikipedia.org/wiki/Double_factorial#Generalizations) 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

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So we got 100 using all the i's from 1 to 9.

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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 $.i$