The Devil is in the Details

Because the limit is in the form of $\frac {0}{0}$ at $x = 1$ , so L'hopital rule is applicable

Which means we need to evaluate the first derivative of $f_{666} (x)$ at $x=1$

Claim : $f_n ' (1) = 1$ for all positive integers $n$ .

Proof by Mathematical Induction :

Basic Step: for $n = 1$ , $f_1 ' (1) = 1$ , which is true

Inductive Step: for a positive integer $k$ , we assume $f_k ' (1) = 1$

By recurrence relation, $\ln \left ( f_{k+1} (x) \right ) = f_k (x) \cdot \ln (x)$

Differentiation gives $\large \frac { f_{k+1} ' (x) }{f_{k+1} (x)} = \frac {f_k (x)}{x} + f_k '(x) \cdot \ln (x)$

Substitution of $x = 1$ proved that it's true for $n=k+1$ too.

Thus the limit is equals to $\large \frac { f_{666} '(1) }{666 \cdot 1^{665} } = \frac {1}{666} \Rightarrow A+B = 667 = 23 \times 29 \Rightarrow D = 29 - 23 = 6$

Hence, with the knowledge of $\displaystyle \sum_{i=1}^n i = \frac {n}{2} (n+1)$ . In this case $n = D^2 = 36$ yields the answer $\boxed{666}$

$As\quad { f }_{ n+1 }(x)={ x }^{ { f }_{ n }(x) }\quad and\quad { f }_{ 0 }(x)=1\quad so,\\ \\ { f }_{ 1 }(x)={ x }^{ 1 },{ f }_{ 2 }(x)={ x }^{ x },{ f }_{ 3 }(x)={ x }^{ { x }^{ x } }\quad and\quad so\quad on\\ \\ { f }_{ 1 }^{ ' }(x)=1,\\ \\ { f }_{ n+1 }^{ ' }(x)={ f }_{ n }(x).{ f }_{ n }^{ ' }(x).{ x }^{ { f }_{ n }(x)-1 }\\ \\ or,\quad \frac { { f }_{ n+1 }^{ ' }(x) }{ { f }_{ n }^{ ' }(x) } =\cfrac { { f }_{ n+1 }(x).{ f }_{ n }(x) }{ x } \quad \quad \quad \quad \quad \quad \quad \quad ......(i)\\ \\ As\quad x\rightarrow 1\\ \\ { f }_{ 1 }(x)\rightarrow 1\\ \\ { { f }_{ 2 }(x)\rightarrow 1 }\\ \\ ..............\\ \\ { f }_{ n }(x)\rightarrow 1\\ \\ so,\\ \\ \cfrac { A }{ B } ={ lim }_{ x\rightarrow 1 }\cfrac { { f }_{ 666 }(x)-1 }{ { x }^{ 666 }-1 } ={ lim }_{ x\rightarrow 1 }\cfrac { { f }_{ 666 }^{ ' }(x) }{ { 666.x }^{ 665 } } \quad \quad .....(ii)\\ \\ by\quad { eq }^{ n }\quad (i),\\ \\ { lim }_{ x\rightarrow 1 }\frac { { f }_{ n+1 }^{ ' }(x) }{ { f }_{ n }^{ ' }(x) } =1\\ \\ \Rightarrow { lim }_{ x\rightarrow 1 }{ f }_{ 666 }^{ ' }(x)={ lim }_{ x\rightarrow 1 }{ f }_{ 665 }^{ ' }(x){ =lim }_{ x\rightarrow 1 }{ f }_{ 664 }^{ ' }(x)\\ \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad ={ .......=lim }_{ x\rightarrow 1 }{ f }_{ 1 }^{ ' }(x)=1\\ \\ by\quad { eq }^{ n }\quad (ii),\\ \\ \cfrac { A }{ B } =\cfrac { 1 }{ 666 } \quad \Rightarrow \quad A+B=667=23\times 29\quad \Rightarrow \quad D=29-23=6\\ \\ Ans=\cfrac { ({ D }^{ 2 })({ D }^{ 2 }+1) }{ 2 } =666$