Now it comes 11

We will show that if $n\equiv11 (\mod100)$ , then $11^n\equiv11 (\mod100)$ . Indeed, by the binomial theorem, $11^n=(1+10)^n\equiv1+10n\equiv11(\mod 100)$ . This shows, by induction, that the numbers $11,11^{11},11^{11^{11}},11^{11^{11^{11}}},...$ all end with the digits 11.

The powers of 11 mod 10 go like this: 01, 11, 21, 31, 41, etc. Thus $11^{11}$ is the same as 1 mod 10. We have just figured out what that produces, so $11^1=\boxed{11}$ .

Finding the last two digits of a number is equivalent to working modulo $100$ . By Euler's theorem, we only need to determine the value of the exponent ${11^{11}}$ modulo $\phi(100)=100\times\left(1-\frac {1}{2}\right)\times\left(1-\frac {1}{5}\right) = 40.$ $11^{11} \equiv \left(121\right)^5\times{11} \equiv11 \pmod{40}$ Now, we need to determine the value of $11^{11}$ modulo $100$ . Then $\begin{array}{c}&11^{11}&\equiv \left(121\right)^5\times{11}\\&\equiv\left(41\right)^2\times{21} \times{11} \equiv81\times{21}\times{11}\\&\equiv18711\equiv11 \pmod{100} \end{array}$