Seeing if the multiplicative group $GF(2^{2016})$ has a subgroup of order 2016. How can I solve this.
I want to find out if the multiplicative geoup $GF(2^{2016})$ has a subgroup of order 2016. For this I need to check if 2016 divides $2^{2016}$. 
Is there an easier way to see this without computing with such large numbers?
Thanks in advance
I have changed the title of this question to reduce confusion.
Thank you all for your help.
 A: Certainly $2016$ does not divide $2^{2016}-1$.
If, more generally, you want to find the exact class of $2^{2016}$ in ${\Bbb Z}_{2016}$ since Euler's theorem does not apply you may proceed as follows.
First observe that
$$
2^{11}=2048\equiv2^5=32\bmod2016.
$$
Then use this congruence to lower the exponent of $2$. Since $2016=11^3+5\cdot11^2+7\cdot11+3$ you get first
$$
2^{2016}=2^{11^3}\cdot(2^{11^2})^5\cdot(2^{11})^7\cdot2^3\equiv
2^{5^3}\cdot2^{5^3}\cdot2^{10}\cdot2^3\bmod2016
$$
(this is justified by $2^{ab}=(2^a)^b$) and then you proceed again "substituting" every $2^{11}$ in the congruence with a $2^5$ until the final exponent becomes feasable.
A: $2016$ clearly doesn't divide $2^{2016}$ -- the only numbers that divide a power of $2$ are other (smaller) powers of $2$, and $2016$ is not a power of $2$ (it is strictly between $2^{10}=1024$ and $2^{11}=2048$).
However, as user438576 points oout,  that is not the right question to ask.
You should be asking whether $2016$ divides $2^{2016}-1$, because that is the order of the multiplicative group of $GF(2^{2016})$ (which doesn't contain the zero of the field).
When does an even number divide an odd number?
A: I see now. First of all the multiplicative group $GF(2^{2016})$ has $2^{2016} - 1$ elements, since zero is not in it. It is now easy to see that 2016 can't divide $2^{2016}$, hence there's no subgroup of order 2016 for this group.
I have changed the title of this question to reduce confusion.
Thank you all for your help.
