Small Representations of $2016$ It's the new year at least in my timezone, and to welcome it in, I ask for small representations of the number $2016$.
Rules: Choose a single decimal digit ($1,2,\dots,9$), and use this chosen digit, combined with operations, to create an expression equal to $2016$. Each symbol counts as a point, and the goal is to minimize the total number of points.
Example (my best so far):
$$2016=\frac{(4+4)!}{4!-4}$$
This expression scores 11 points. That is 2 points for the parentheses, 4 points for the $4$s, 1 point for $+$, 2 points for the $!$s, 1 point for the fraction, and 1 point for the $-$.
Allowable actions: basic arithmetic operations (addition, subtraction, multiplication, division), exponentiation, factorials, repeated digits (i.e. if you are working with the digit $7$, you can use $77$ for 2 points), and use of parentheses.

What is the minimum number of points for an expression of the above form equaling 2016, and what are those minimum expressions?

Note that by "Use a single decimal digit" I mean you may only use one of the digits $1$ through $9$, so for example, you can't save in the above expression just by using $8!$ instead of $(4+4)!$ because you would still have the $4!-4$ part.
This question is mostly for fun, but could have some relevance to students who participate in thematic math competitions this year.
 A: $2016 = 2^{\,22\,/\,2} - 2\;2^{\,2^{\,2}}\;$ with just elementary operations (10 symbols).

[EDIT]  P.S.  A few more variations with and without the × disputed in the comments.
$$
2016 = 2^{\,22\,/\,2} - 2^{\,2^{\,2}}\;2 = 2^{\,22\,/\,2} - 2 \cdot 2^{\,2^{\,2}} = \sqrt 2 ^{\,22} - 2 \; 2^{\,2^{\,2}} = \sqrt 2 ^{\,22} - 2 \cdot 2^{\,2^{\,2}}
$$
A: Since $2016 = \# {\rm GL}_2({\bf Z} / 7 {\bf Z})$
[alas I don't remember from whom I learned this a year or so ago],
we can even use $7$'s nicely, though not enough to win this "contest",
e.g. $(7 \times 7 - 7) \times (7 \times 7 - \frac77)$.
A: 9 symbols using, appropriately, the digit 9:
$$
\frac{9!}{99 + 9 \times 9}
$$
(I thought $6.6 \times 6! - 6!$ was another 9-symbol solution
if decimal points are allowed, but it actually gives $4032$
which is exactly twice too big.)
[added later: then there's the puzzle of writing 201 (not 2016) using
four 9's and any number of the standard operations $+$ $-$ $\times$ $\div$ ! $\sqrt{\phantom0}$ and parentheses.]
A: Make sure you follow your New Year's resolution $24$ hours a day, $7$ days a week, for $12$ months $24 \cdot 7 \cdot 12 = 2016$.
A: Here's a devil of an answer:
$$2016=666+666+666+6+6+6$$
A: With binomial coefficients ($8$ symbols):
$$2016={64 \choose 2}={2^{2^2+2}  \choose 2}$$
It can be expected that many olympiad problems in $2016$ will use this combinatorial property.
P.S. Special thanks to Alex Fok for minus one symbol in $64$.
A: $$\sum_3^{3\times3}n^3$$ is 7 points, right?
A: How about $2016=4\sqrt{4}(4^4-4)$ using $9$ symbols.

Edit: Here's another nine:
$2016=3!(333+3)$
A: The first $5$ answer (25 points, lousy)
$55 \frac {55} 5 \frac {5!!} 5 + 5 (55 - 5!!) + \frac 5 5$
Didn't see a $5$ answer yet. Threw my hand in the ring for some digit representation.
A: 13 points using $2$s
$$ 2016 = 2^{2^2} ( 2^{2^2} 2^2 2 - 2) $$
19 points using $6$s
$$ 2016 = 6! + 6! + 6! - (6+6)(6+6) $$
Many points using $8$s (involving $8^{5/3}=32$)
$$ 2016 = \left( \frac{8 \cdot 8}{88 - 8 \cdot 8} - \frac{8}{8} \right) \left(8 \cdot 8 - \frac{8}{8} \right) $$
