In addition to the calculation implied by Timothy's answer, let me describe a way to see it directly. Consider $\sum_{r=0}^{n-1}C(n-2,r)$. This is the sum of
(the number of ways of choosing a subset of $0$ elements out of a set of $n-2$ elements)
+ (the number of ways of choosing a subset of $1$ element out of a set of $n-2$ elements)
+ (the number of ways of choosing a subset of $2$ elements out of a set of $n-2$ elements)
. . .
+ (the number of ways of choosing a subset of $n-1$ elements out of a set of $n-2$ elements)
Since any subset of $n-2$ elements has either 0, 1, 2, . . . , or $n-2$ elements, the total is just the number of ways of choosing any subset of $n-2$ elements. And this is just $2^{n-2}$.