Why do remainders show cyclic pattern? Let us find the remainders of $\dfrac{6^n}{7}$,
Remainder of $6^0/7 = 1$
Remainder of $6/7 = 6$
Remainder of $36/7 = 1$
Remainder of $216/7 = 6$
Remainder of $1296/7 = 1$
This pattern of $1,6,1,6...$ keeps on repeating. Why is it so? I'm asking in general, that is for every case of type $a^n/b$'s remainder keeps on repeating as we increase $n$.
P.S: This is a follow up question of my previous question.

Using your example, there are only $6$ possible remainders. I am excluding $0$ because $6^n$ will never be a multiple of $7$.
So you start with $6^0 = 1$.
Then $6^1 \equiv (6^0)\times 6 \equiv 1\times 6 \equiv 6 \pmod 7$
Then $6^2 \equiv (6^1)\times 6 \equiv 6\times 6 \equiv 1 \pmod 7$
And we are back to $1$ and we are now forced into a repeating pattern. Since there are only $6$ possible remainders, we are going to have to repeat a remainder by at most the $6$'th computation. From that point on, we must have a repeating pattern.
