Formulae of the Year $2016$

Question

Soon it's the year $2016$. Time to ponder how we can arrange the digits in 2016 to form a valid equation. Use any symbols you like (please explain the less obvious ones). Keep digits in the same order (should this be relaxed?).

Examples:

$$\lfloor e^2\rfloor + 0 - 1! = 6$$ $$\left\lfloor\sqrt{\sqrt{201}}\right\rfloor = \lceil\sqrt{6}\rceil$$

where $\lfloor x\rfloor$ denotes the floor function and $\lceil x\rceil$ the ceiling.

Don't overuse constants (i.e. avoid adding up several $\pi$ and $e$ just to get to some arbitrary value).

EDIT: clarification: use each of the digits $2$, $0$, $1$, $6$ in this order only once. Combine digits giving $20$, $201$, $16$, etc as you like (I won't argue whether in a fraction the numerator or denominator comes first :-). Please don't criticize answers that violate this rule, as this clarification came late.

Answer 1

An easy one ;-) $$(2 + 0 + 1)! = 6$$

Answer 2

Arithmetic's fundamental theorem implies $$2016=2^{5}\cdot3^{2}\cdot 7$$

Edit:

According to Dan Brumleve $$2016=2^{0-1+6}\cdot\left(2^{0+1\cdot6}-2^{0}\cdot1^{6}\right)$$

Answer 3

$$\color{red}{2}\pi i\left(\oint_{|z-\color{red}{0}|=R}\frac{dz}{z}\right)^{-\color{red}{1}}=\lceil \cos(\color{red}{6})\rceil$$

Answer 4

A simple one: $$2\times 0 = \sin\left(1 \times 6 \times \pi \right) $$

Answer 5

Another easy one) $$\Large(\color{red}{ 2}!)^{\color{blue}{\Large {2}}}+(\Large \color{red}{0}!)^{\color{blue}{\Large{0}}}+(\color{red}{1}!)^{\color{blue}{\Large{1}}}=\color{red }{\Large 6}$$