a is a nonzero integer. Is a^a greater than 1?

Target question: Is a^a greater than 1?

Given: a is a nonzero integer.

Statement 1: a < -1
Let's start TESTING some values of a and see if we discover a PATTERN

a = -2. Here a^a = (-2)^(-2) = 1/(-2)^2 = 1/4
a = -3. Here a^a = (-3)^(-3) = 1/(-3)^3 = 1/(-27) = - 1/27
a = -4. Here a^a = (-4)^(-4) = 1/(-4)^4 = 1/256
a = -5. Here a^a = (-5)^(-5) = 1/(-5)^5 = 1/(some negative value) = something negative
a = -6. Here a^a = (-6)^(-6) = 1/(-6)^6 = 1/(some positive integer) = a positive fraction that's less than 1
a = -7. Here a^a = (-7)^(-7) = 1/(-7)^7 = 1/(some negative value) = something negative
.
.
.
As we can see, when a is a NEGATIVE EVEN number, a^a = some positive fraction that's LESS THAN 1
When a is a NEGATIVE ODD number, a^a = some negative value that's LESS THAN 1
So, in both possible cases, a^a is definitely LESS THAN 1
Since we can answer the target question with certainty, statement 1 is SUFFICIENT

Statement 2: a is even
We already saw in statement 1 that if a is a NEGATIVE EVEN integer, then a^a = some positive fraction that's LESS THAN 1
What if a is a POSITIVE EVEN integer?
Let's test some possible values:
a = 2. Here a^a = (2)^(2) = 4. Here, a^a is GREATER THAN 1
So, we'll stop right here.
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

Answer: A

Cheers,
Brent