2014^2014

I'm sure there's an easier way to do this, but when in doubt, use exponentiation by squaring:

$\begin{aligned} 14^{2014}&\equiv196^{1007}\\&\equiv196^{1006}(196)\\&\equiv416^{503}(196)\\&\equiv416^{502}(536)\\&\equiv56^{251}(536)\\&\equiv56^{250}(16)\\&\equiv136^{125}(16)\\&\equiv136^{124}(176)\\&\equiv496^{62}(176)\\&\equiv16^{31}(176)\\&\equiv16^{30}(816)\\&\equiv256^{15}(816)\\&\equiv256^{14}(896)\\&\equiv536^7(896)\\&\equiv536^6(256)\\&\equiv296^3(256)\\&\equiv296^2(776)\\&\equiv616(776)\\&\equiv\boxed{16}\pmod{1000} \end{aligned}$

$2014^{2014}\equiv 14^{2014}\pmod{1000}$

By Euler's Theorem:

$\left(14^{\phi(1000)}\right)^5=\left(14^{400}\right)^5=14^{2000}\equiv 1^5=1 \pmod{1000}$

So: $14^{2014}= 14^{2000}\cdot 14^{14}\equiv 14^{14} =2^{14}\cdot 7^{14}\pmod{1000}$

I don't know a short-cut for the rest calculation. However, if one doesn't want to touch calculator, then he can work out the rest step by step multiplying and then taking modulo $1000$ :

$2^{14}=2^{10}\cdot 2^4=1024\cdot 16\equiv 24\cdot 16 =384 \pmod{1000}$

$7^{14}=2401\cdot 7^{10}\equiv 401\cdot 7 \cdot 7^9\equiv 807\cdot 7^9=\cdot\cdot\cdot=407\cdot 7\equiv 849 \pmod{1000}$

$\therefore ~~~ 2014^{2014}\equiv 14^{14}=2^{14}\cdot 7^{14}\equiv 384\cdot 849 = 326016\equiv \fbox{016}\pmod{1000}$

Okay, I solved this problem in a real lengthy method. I'm sure there may be an easier approach, so if any of you guys get to know any shorter method which will save a lot of time, do tell me. The remainder when 2014^2014 is divided by 1000 is actually the last three digits of the number 2014^2014. So using modulo function, we'll see that 2014^2000 will leave a remainder 376 when divided by 1000. Moreover, 2014^14 will leave a remainder 16 when divided by 1000. So the remainder left by 2014^2014 when divided by 1000 will be 16. Actually the last three digits are 016, but we neglect 0 here because they have asked for the remainder.