The trend of huge number

$\begin{cases} 2016 & \equiv 16 \text{(mod 100)} (5n+1) \\ 2016^{2} & \equiv 56 \text{(mod 100)} (5n+2) \\ 2016^{3} & \equiv 96 \text{(mod 100)} (5n+3) \\ 2016^{4} & \equiv 36 \text{(mod 100)} (5n+4) \\ 2016^{5} & \equiv 76 \text{(mod 100)} (5n) \\ \end{cases}$

$\begin{cases} 2^{5} & \equiv 2 \text{(mod 5)} (4n+1) \\ 2^{6} & \equiv 4 \text{(mod 5)} (4n+2) \\ 2^{7} & \equiv 3 \text{(mod 5)} (4n+3) \\ 2^{8} & \equiv 1 \text{(mod 5)} (4n) \end{cases}$

$\begin{cases} 4^{3} & \equiv 4 \text{(mod 5)} (2n+1) \\ 4^{4} & \equiv 1 \text{(mod 5)} (2n) \end{cases}$

$2017 \equiv 2 \text{(mod 5)}$

${2017}^{2018} \equiv 2^{2018} \text{(mod 5)}$

${2017}^{2018} \equiv 4 \text{(mod 5)}$

${2017}^{{2018}^{2019}} \equiv 4^{2019} \text{(mod 5)}$

${2017}^{{2018}^{2019}} \equiv 4 \text{(mod 5)}$

${2017}^{{2018}^{{2019}^{2020}}} \equiv 4^{2020} \text{(mod 5)}$

${2017}^{{2018}^{{2019}^{2020}}} \equiv 1 \text{(mod 5)}$

${2017}^{{2018}^{{2019}^{{2020}^{20162016}}}} \equiv 1^{20162016} \text{(mod 5)}$

${2017}^{{2018}^{{2019}^{{2020}^{20162016}}}} \equiv 1 \text{(mod 5)}$

$2016^{{2017}^{{2018}^{{2019}^{{2020}^{20162016}}}}}=2016^{5m+1}$ ,where $m$ is a positive integer.

$\Rightarrow2016^{{2017}^{{2018}^{{2019}^{{2020}^{20162016}}}}} \equiv 16 \text{(mod 100)}$

By Chinese remainder theorem $\pmod{100}$ is the same as $\pmod{25}$ and $\pmod{4}$ .

---

Since $2016 = 4 * 504$ we know that the number is $\equiv 0 \pmod{4}$

---

$\pmod{25}$

$25$ and $16$ are coprime, so we can use Euler's totient function $\phi{25}=20$ .

$20$ and $17$ are coprime, so, again, we can use Euler's totient function $\phi{20}=2^3$ .

$2018^{2019^{2020^{20162016}}} \equiv 2^{2019^{2020^{20162016}}} \equiv 0 \pmod{2^3}$ .

Now we have much simpler expression:

$16^{17^{0}} =16 \pmod{25}$

---

Since $x \equiv 0 \pmod{4}$ and $x \equiv 16\pmod{25}$ we notice that $\boxed{16}$ is the answer.

I used the fact that $(\text{even no.})^{20n}\equiv76\pmod{100}$ where $\text{n}$ is any natural number and the number is not an even multiple of $5$ .

So we need to find $2017^{2018^{2019^{2020^{20162016}}}}\pmod{20}$

---

$2017^{2018^{2019^{2020^{20162016}}}}\equiv17^{2018^{2019^{2020^{20162016}}}}\equiv(-3)^{2018^{2019^{2020^{20162016}}}}\equiv3^{2018^{2019^{2020^{20162016}}}}\pmod{20}$

Now, we know that $3^{4}\equiv1\pmod{20}$ and $2018^{a}$ where $\boxed{a > 2}$ is divisible by $4$ .

Therefore,

$3^{2018^{2019^{2020^{20162016}}}}\equiv3^{4k}\equiv(3^{4})^{k}\equiv1^{k}\equiv1\pmod{20}$

So, $2017^{2018^{2019^{2020^{20162016}}}}\equiv1\pmod{20}$ .

Hence, $2016^{2017^{2018^{2019^{2020^{20162016}}}}}\equiv2016^{20n + 1}\equiv(2016^{20})^{n}\times2016\equiv76^{n}\times2016\equiv76\times16\equiv{16}\pmod{100}$ ...because $76$ raised to the power any natural number gives $76$ as the last two digits.

So our answer becomes $\boxed{16}$ .

---

$\underline{NOTE :}$

By Euler's Theorem - $n^{\phi(25)}\equiv1\pmod{25}$ where $\text{n}$ is co - prime to $25$ .

So $(\text{even no.})^{20}\equiv1\pmod{25}$ . Hence possible last two digits are $01 , 26 , 51 , 76$ .............. $\boxed{\phi(25) = 20}$

But $01$ and $51$ are not possible since even number raised to the power anything is even.

$26$ is also not possible because even number raised to the power $20$ is obviously divisible by $4$ .

Hence possible last two digits is $\boxed{76}$ only.