Last Digits - 4

$2^{\phi(25)}=2^{20}=16^5\equiv 1 \pmod{25}$ so $16^{2015}\equiv 1 \pmod{25}$ and $16^{2016}\equiv \boxed{16} \pmod{100}$

See, I'll go as easy as I can, I am only going to write the last two digits of the expansion of 2016 till (2016)^6 (2016)^1= 16 (2016)^2= 56 (2016)^3= 96 (2016)^4= 36 (2016)^5= 76 (2016)^6= 16 ................ You can very well get the intuition the each digit is as if 40 more than the previous one even as if during the 4th expansion (96+40=136), where we take the last two digits 36, also as if during the 6th expansion (76+40=116) where we take the last two digits 16 (Just take it in that sense!). This goes on. Now see, What is true for (2016)^6 is true for (2016)^11, 16, 21 and so on(Not just 6, every interger in power of 2016 gives the same last digit after having a +5 increment(An Arithmetic progression of 5 to say the least), visible from the pattern above), now the number on the power of the required question i.e 2016 only happens when only number 6 gets an increment of 5 for the 403rd time( No other number from 1-9 can after an repeated progression of +5 become 2016 except number 6 respectively), Again since 2016^6= last digit 16 it follows 2016^2016=.....16 as well, Please pardon me if I failed to be more logical and if you find a better explanation, kindly post here....