For any positive integer $k$, let $f_1(k)$ denote the square of the sum of the digits of $k$.  For $n \ge 2$, let $f_n(k) = f_1(f_{n - 1}(k))$.  Find $f_{1988}(11)$.
We see that $f_{1}(11)=4$
$f_2(11) = f_1(4)=16$
$f_3(11) = f_1(16)=49$
$f_4(11) = f_1(49)=169$
$f_5(11) = f_1(169)=256$
$f_6(11) = f_1(256)=169$
Note that this revolves between the two numbers. Since $1988$ is even, we thus have $f_{1988}(11) = f_{4}(11) = \boxed{169}$.