A sequence of polynomials

The trick is to express each term in terms of $2$ raised to the power of a power of $2$ , e.g. $256 = 2^{2^3}$

$\begin{aligned} \LARGE f_{100} (2) & = & 2^{2^{100-99-1}} + f_{99} \left ( 2^{2^{100-99}} \right ) \\ \LARGE & = & 2^{2^{100-98-1}} + 2^{2^{100-98-2}} + f_{98} \left ( 2^{2^{100-98}} \right ) \\ \LARGE & = & 2^{2^{100-97-1}} + 2^{2^{100-97-2}} + 2^{2^{100-97-3}} + f_{97} \left ( 2^{2^{100-97}} \right ) \\ \LARGE & = & \ldots \\ \LARGE & = & 2^{2^{100-1-1}} + 2^{2^{100-1-2}} + 2^{2^{100-1-3}} + \ldots + 2^{2^{100-1-99}} + f_{1} \left ( 2^{2^{100-1}} \right ) \\ \LARGE & = & 1 + \displaystyle \sum_{k=0}^{98} 2^{2^k} \\ \LARGE & = & 1 + 2^{2^0} + 2^{2^1} + \displaystyle \sum_{k=2}^{98} 2^{2^k} \\ \LARGE & = & 7 + \displaystyle \sum_{k=2}^{98} 2^{2^k} \\ \end{aligned}$

Notice that $2^{2^4} = 16$ , since we only consider the last two digits, that is, modulo $100$ , we have $16^2 \equiv 56$ , $56^2 \equiv 36$ , $36^2 \equiv 96 \equiv -4$ , $(-4)^2 \equiv 16$ which indicates the last two digits repeats after a cycle of $4$

So for $k = 2,3,4, \ldots, 98$ , we consider splitting them into sets 4 sets as such

$\begin{aligned} S_1 & = & \{ 2, 6, 10, 14, \ldots 94, 98 \} \Rightarrow \text{ a total of } 25 \text { elements} \\ S_2 & = & \{ 3, 7, 11, 15, \ldots 95 \} \Rightarrow \text{ a total of } 24 \text { elements} \\ S_3 & = & \{ 4, 8, 12, 16, \ldots 96 \} \Rightarrow \text{ a total of } 24 \text { elements} \\ S_4 & = & \{ 5, 9, 13, 17, \ldots 97 \} \Rightarrow \text{ a total of } 24 \text { elements} \\ \end{aligned}$

Continue from above

$\begin{aligned} \LARGE f_{100} (2) & \equiv & 7 + \sum_{k \in S_1} 2^{2^k} + \sum_{k \in S_2} 2^{2^k} + \sum_{k \in S_3} 2^{2^k} + \sum_{k \in S_4} 2^{2^k} \\ \LARGE & \equiv & 7 + 25(16) + 24(56) + 24(36) + 24(-4) \\ \LARGE & \equiv & 7 + 24(16 + 56 + 36 - 4) + 16 \\ \LARGE & \equiv & \boxed{19} \\ \end{aligned}$