mathematics - 6, the magic number

Here's a fun (albeit difficult) one:

Make these equations true using arithmetic operations:

1 1 1 = 6
2 2 2 = 6
3 3 3 = 6
4 4 4 = 6
5 5 5 = 6
6 6 6 = 6
7 7 7 = 6

8 8 8 = 6
9 9 9 = 6

For example: 6 + 6 - 6 = 6 (I hope I did not spoil some of you :D)

Allowed operators are:

+, -, *, /, ! , ^, %

Setting parenthesis is also allowed.

The ^ operator is an exception as you are permitted to supply a second argument to it which may be any positive integer or the multiplicative inverse of it.

$x^{1/y}$ is always positive and real.

If you find an alternative solution using other operators you may post it but please also provide a solution using only these 7 operators.

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For those of you who think this was easy, here is a bonus:

0 0 0 = 6

Answer

1.

$(1+1+1)! = 6 $

2.

$2+2+2 = 6$

3.

$3*3-3 = 6$

4.

$\left(4-\frac 4 4\right)! = \sqrt 4+\sqrt 4+\sqrt 4=6$

5.

$5+\frac 5 5 = 6$

6.

$6*\frac 6 6 = 6 + 6 -6=6$

7.

$7-\frac 7 7 = 6$

8.

$\left(\sqrt{8+\frac 8 8}\right)! = 6$

9.

$\left(\frac{\sqrt{9}\sqrt{9}}{\sqrt 9}\right)! = 6$

Bonus:

$(0!+0!+0!)! = 6$