Last two digits here, is impractical part 1.

SOLUTION:

If you observe then, you will find that as $\displaystyle 11^2$ ends with $21$ , $\ \ \displaystyle 11^9$ will end with $91$ .

Thus $\displaystyle 11^{10} \equiv 1 \pmod {100} \longrightarrow 11^{11} \equiv 11 \pmod {100}$ .

Hence $\displaystyle 11^{11^{11^{^{ .^ {.^ .}}}}} \bmod {100} \ \ \ \$ would reduce to $\ \ \ \ \displaystyle 11^{11} \bmod {100}$ .

Thus $\displaystyle 11^{11^{11^{^{ .^ {.^ .}}}}} \equiv 11 \pmod {100}$ .

Let $n = 11^{11^{11^{\cdot^{\cdot^\cdot}}}}$ . Then

$\begin{aligned} 11^{11^{11^{\cdot^{\cdot^\cdot}}}} & \equiv 11^n \text{ (mod 100)} \\ & = (10+1)^n \text{ (mod 100)} \\ & = \left(10^n + 10^{n-1}+ \cdots + \frac {100n(n-1)}2 + 10n + 1 \right) \text{ (mod 100)} \\ & = \left(0 + 0+ \cdots + 0 + 10\blue n + 1 \right) \text{ (mod 100)} & \small \blue{\text{Note that }n \text{ ends with 1.}} \\ & = 10 + 1 \equiv \boxed {11} \text{ (mod 100)} \end{aligned}$

Let $x = {11}^{{11}^{...}} \mod 100$ . Then since 100 and 11 are relative primes ${11}^{100} = 1 \mod 100$ . That means $x = {11}^{x} \mod 100$ . The obvious and only value for x is $\boxed{11}$