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.
 A: The answer is
$$ 6 = -1 \pmod 7. $$
Therefore, you have $(-1)^{n}$ and it depends only on $n$ and it $-1$ or $1$.
More detailed theory you can find following this link: Modular arithmetic
P.S. $ number \pmod 7 = remainder$
A: Note that as you find more and more powers, you'll eventually run out of remainders. So you'll come back to a previous number, and since exponentiation is repeated multiplication, you'll generate the cycle again. As an example, let's look at powers of $2$ modulo $7$. We have
\begin{align*}
2^1 &\equiv 2 \pmod{7} \\
2^2 &\equiv 2\cdot 2 \equiv 4 \pmod{7} \\
2^3 &\equiv 4 \cdot 2 \equiv 1 \pmod{7} \\
2^4 &\equiv 1 \cdot 2 \equiv 2 \pmod{7} \\
2^5 &\equiv 2 \cdot 2 \equiv 4 \pmod{7} \\
2^6 &\equiv 4 \cdot 2 \equiv 1 \pmod{7} 
\end{align*}
Hopefully, this clarifies how multiplication works in modular arithmetic too.
A: The reason is simple: a remainder can be expressed as a function of the previous remainder alone.
Indeed, let $r_n=6^n\bmod 7$, i.e. $6^n=7q_n+r_n$.
Then $\color{blue}{r_{n+1}}=6^{n+1}\bmod 7=(6\cdot6^n)\bmod 7=(42q_n+6r_n)\bmod 7=\color{blue}{(6r_n)\bmod 7}.$
As there is a finite number of possible remainders, the sequence must be periodic.
$$\begin{align}1&&\to1\\6&\to6\cdot1&\to6\\36&\to6\cdot6&\to1\\216&\to6\cdot1&\to6\\1296&\to6\cdot6&\to1\\\cdots\end{align}$$
or if you prefer
$$\begin{align}1&\to1\\6\cdot1&\to6\\6\cdot6&\to1\\6\cdot1&\to6\\6\cdot6&\to1\\\cdots\end{align}$$
A: First, let me address a simpler question: why is there a pattern to the remainders at all?
Suppose I have any positive integer $b$ (in your case $b=7$). Then remainders respect multiplication: if $x, x'$ are positive integers which yield the same remainder when divded by $b$, and $y, y'$ are positive integers which yield the same remainder when divded by $b$, then $xy$ and $x'y'$ yield the same remainder when divided by $b$.
This is modular arithmetic: for a positive integer $b$, we can develop arithmetic on the set $R_b=\{0, 1, 2, . . . , b-1\}$ by interpreting elements of $R_b$ as remainders. For instance, any time I multiply two even numbers I get an even number; this is represented as $$0\times 0=0 \quad \mbox{(in $R_2$)}.$$ Because remainders respect multiplication, this makes sense and works the way we want it to.
So what does this have to do with powers? Well, fix $b$, and consider the sequence $a, a^2, a^3, . . .$. Since $R_b$ is finite there are some $m<n$ such that $a^m$ and $a^n$ leave the same remainder when divided by $b$. Once this happens, the remainders will repeat: the remainder of $a^{n+1}$ when divided by $b$ will be the same as the remainder of $a^{m+1}$ when divided by $b$, etc. So the sequence of remainders of powers of some number will always be eventually repeating.
Note that "eventually" is crucial here: consider $a=3$, $b=27$. Then the sequence of remainders is $3, 9, 0, 0, 0, 0, . . .$. Similarly, if we take $a=2$ and $b=12$ we get $2, 4, 8, 4, 8, 4, 8, . . .$ This happens because $a$ and $b$ share prime factors.  Euler's theorem says that if, by contrast, $a$ and $b$ are coprime (have no factors besides $1$ in common), then this pattern is just repeating, full stop, and in fact tells us how long it will take to repeat.
A: Looking at remainders after division by 7 is called arithmetic modulo 7.
You are regarding powers of 6, modulo 7. But $6$ is $-1$, modulo $7$. This is written:
$$6 \equiv -1 \pmod 7$$
But the powers of $-1$, that is the numbers $(-1)^n$, simply alternate: $1,-1,1,-1,1,\ldots$.
A: 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.
