Two keys here:
If 1 > a > 0, then 1 > a² > a� > 0.
If a > 1, then a� > a² > 1.
S1 tells us that a� > 1. This only works in the second case above, so a² > 1; SUFFICIENT.
S2 is easier if we raise both sides to the sixth power. Then we have a³ > a². This is the same as the case above. (If a > 1, then of course a³ > a² > 1.) So this is also SUFFICIENT.
AS #50
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