Waiting for 2016!

Let $N=\large {\color{#3D99F6}{2016}^{\color{#20A900}{2016}^{\color{#D61F06}{2016}^{\color{#624F41}{2016}^\color{magenta}{2016}}}}}$ .

Finding the last two digits of a number is equivalent to determining the remind of a number modulo $100$ . Since $\gcd\left(2016,100\right)\neq1$ , we can't use Euler's theorem directly. Instead, we need to determine the remind of $N$ modulo $4$ and modulo $25$ .

Due to $2016$ is a multiple of $4$ , according to the rules of modular arithmetic, hence ${2016}^{n}\equiv0\pmod{4}\text{ for }n\geq1$ . Furthermore, we will find the residue of $N$ modulo $25$ . Note that $\gcd\left(2016, 25\right)=1$ , so, we can use Euler's theorem. To avoid confusion, we will use the original question.

$\begin{aligned}\large{\color{#624F41}{2016}^\color{magenta}{2016}}&\equiv0\pmod{4}\\ \large{\color{#D61F06}{2016}^{\color{#624F41}{2016}^\color{magenta}{2016}}}&\equiv\color{#D61F06}{2016}^{0}\equiv1\pmod{8}\\ \large{\color{#20A900}{2016}^{\color{#D61F06}{2016}^{\color{#624F41}{2016}^\color{magenta}{2016}}}}&\equiv\color{#20A900}{2016}^{1}\equiv16\pmod{20}\\ \large {\color{#3D99F6}{2016}^{\color{#20A900}{2016}^{\color{#D61F06}{2016}^{\color{#624F41}{2016}^\color{magenta}{2016}}}}}&\equiv\color{#3D99F6}{2016}^{16}\equiv{16}^{16}\equiv{256}^{8}\equiv{6}^{8}\equiv{1296}^2\equiv{\left(-4\right)^2}\equiv16\pmod{25}\end{aligned}$

Now, we have: $\large {\color{#3D99F6}{2016}^{\color{#20A900}{2016}^{\color{#D61F06}{2016}^{\color{#624F41}{2016}^\color{magenta}{2016}}}}}\equiv\begin{cases}0\pmod{4}\\16\pmod{25}\end{cases}$

Using Chinese Remainder theorem, we will get $\large {\color{#3D99F6}{2016}^{\color{#20A900}{2016}^{\color{#D61F06}{2016}^{\color{#624F41}{2016}^\color{magenta}{2016}}}}}\equiv16\pmod{100}$