prove that 2016 is a self-composable number

Question

Define a number self-composable if it may be computed using just the digits of the number itself (used just once) and the following operations:

• basic operations ( $+, -, \times, \div$)
• less than basic operations ($x^y, \;\sqrt x, \;x!, \;x.y$ (decimal point))
• extended operations ($.x, \; .\overline{x}$)
• parentheses at will

If the digits of the mathematical operation are in the same order as in the number itself, the number is said orderly self-composable.

For example, 25 is self-composable ($5^2 = 25$) and 343 is orderly self-composable, since $(3+4)^3 = 343$.

2016 is self-composable too: find how.

Answer 1

I think I have it

$ 2016 = ({.2} / {.\overline1} + 0!) \times 6!$

Also as pointed out by Matt in the comments, we can swap things around so that 2016 is orderly self-composable

$ 2016 = ({.2} / .\overline{0!} + 1) \times 6!$

Answer 2

find divisors of 2016 that contains its own numbers: $2016/2 \rightarrow 1008/2 \rightarrow 504/2 \rightarrow 252/2 \rightarrow 126/126 \rightarrow 1$

therefore;

we can't use $2^4$ since $4$ is not in $2016$ but we can do $2^2$ for $2$: $2^2 \times 2^2 \times 126 \times 1^0$