Starting with row $ 0 $, the $ k^\text{th} $ row has the numbers \[\binom{k}{0},\binom{k}{1},\binom{k}{2},\ldots,\binom{k}{k} .\]For every number in the row except the first and last values to be even, $ k $ must only have even factors, so it must be a power of $ 2 $. Since the highest power of $ 2 $ under $ 100 $ is $ 2^6=64 $, $ \boxed{6} $ of the first 100 rows have only even numbers besides $ 1 $.