Why does $6^{2m}$ mod 7 $\not\equiv 6^{2n + 1}$ mod 7 for all integers $m$ and $n$?

Question

Other than just observing the pattern that 6 to odd powers mod 7 is 6 and 6 to even powers mod 7 is 1, how would one find out? I assume it has something to do with the prime divisors 2 and 3, but I'm not sure what.

Who's theorems or lemmas should I read?

Answer 1

HINT

Note that

$$6 \equiv -1 \pmod 7$$

and

$$(-1)^{2n}\neq (-1)^{2n+1}$$

Answer 2

Follows from $6\equiv -1 \bmod 7$