For every real number $x \neq - 1$, let $f ( x ) = \frac { x } { x + 1 }$. Write $f _ { 1 } ( x ) = f ( x )$ and for $n \geq 2 , f _ { n } ( x ) = f \left( f _ { n - 1 } ( x ) \right)$. Then,
$$f _ { 1 } ( - 2 ) \cdot f _ { 2 } ( - 2 ) \cdots \cdots f _ { n } ( - 2 )$$
must equal\\
(A) $\frac { 2 ^ { n } } { 1 \cdot 3 \cdot 5 \cdots \cdot ( 2 n - 1 ) }$\\
(B) 1\\
(C) $\frac { 1 } { 2 } \binom { 2 n } { n }$\\
(D) $\binom { 2 n } { n }$.