The remainder when $$((64)^{(64)})^{(64)}$$ is divided by 7 is equal to

First write the given power in a compact form.
$$((64)^{(64)})^{(64)} = \left(64^{64}\right)^{64}$$

Using the law of exponents $$\left(a^{m}\right)^{n}=a^{mn}$$, we get
$$\left(64^{64}\right)^{64}=64^{64\times 64}=64^{4096}$$

We need the remainder of $$64^{4096}$$ when divided by 7. For remainders, replace each base by its least positive residue modulo 7:
$$64 \equiv 64-63 = 1 \pmod{7}$$ because $$7 \times 9 = 63$$.

Therefore
$$64^{4096}\equiv 1^{4096}\pmod{7}$$

Any power of 1 is still 1, so
$$1^{4096}=1$$

Hence the remainder when $$((64)^{(64)})^{(64)}$$ is divided by 7 equals $$1$$.

The correct option is Option B (1).