Patch to the solution for HW 8, problem 5
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This proof actually needs a second base case, for n=3, because the
inductive step is referencing results for the previous two smaller
trees.  The extra base case would say that T_3 consists of a root plus
two edges, so it has height 1, which is equal to n-2.

There may be ways to avoid the second base case by adding some
additional discussion to the inductive step, or by first arguing that
T_k+1 is always at least as tall as T_k.  But we think adding the
second base case is the most elegant method.