Least residue power modulo May someone confirm that the least residue of $44^8$ modulo $7$ is $4$ please?
And that $2^2 \equiv  4\, (\!\!\!\mod 7)$?
Thanks in advance.

It is correct: $44\equiv 2\mod 7$, so  $44^8\equiv 2^8\mod 7$. Now $2^3\equiv 1\mod 7$, so $2^8\equiv 2^{8\bmod 3}=2^2\mod 7=4$, since $0\le 4<7$.
