It's still the early 2015

Firstly, $2^{4x} \equiv 1 \pmod{5}$ with every positive integer $x$ (easily proved).

As a result:

$2^{20}-1=(2^4-1)(2^{16}+2^{12}+2^8+2^4+1)$ is divisible by 25, means $2^{20} \equiv 1 \pmod{25}$ ;

$2^{100}-1=(2^{20}-1)(2^{80}+2^{60}+2^{40}+2^{20}+1)$ is divisible by 125, means $2^{100} \equiv 1 \pmod{125}$

Once more and we have: $2^{500} \equiv 1 \pmod{625}$

So: $2^{2000} \equiv 1 \pmod{625}$

$(2^{2000}-1)$ is divisible by 625 -> $(2^{2004}-2^4)$ is divisible by 10000

Then $(2^{2016}-2^{16})$ is divisible by 10000

The last 4 digits of $2^{2016}$ is the same as that of $2^{16}$ , which is $\boxed{5536}$ .

To get the last 4 digits of $2^{2016}$ , we need to get the remainder when it is divided by 10000.

To get the remainder you need to use CRT or Modular Arithmetic

By Modular Arithmetic:

2^2016mod10000=2^8^252mod10000=256^252mod10000=65536^126mod10000=5536^126mod10000=30647296^63mod10000=7296^63mod10000=(1616^31)(7296)mod10000=(1456^15)(1616)(7296)mod10000=(9936^7)(1456)(1616)(7296)mod10000=(4096^3)(9936)(7296)(2896)mod10000=(7216)(7856)(9216)mod10000=(7216)(896)mod10000=5536mod10000

Therefore, the last 4 digits of $2^{2016}$ is $\boxed{5536 }$

we need to divide by 10,000 and remainder would be last 4 digits. 2^2016=65536^126= (6*10000+5536)^126 = 126C0(10000)126+... last term of expansion would be 5536. Each term has atleast one 10,000. Hence remainder is 5536