Recurring decimal expansion of $\frac17$ The decimal expansion of $\frac17$ seems to have an interesting pattern:
$$ \frac{1}{7} = 0.142857142857142857...$$
Take the first two digits of the expansion: $14 = 2^1*7$
Then the next two: $28 = 2^2*7$
The pattern seems to halt abruptly with the next two: $57 \neq 2^2*7$ (since $2^3*7 = 56$)
But when continued further, the pattern established itself once more, although in a slightly different way: 
$2^4*7 = 112$; the $1$ in the hundreds place appears to have been "carried over".
This leads us to: 
$$\frac 17 = 0.14 + 0.0028 + 0.000056 + 0.00000112 .....$$
Or more concisely, 
$$\frac 17 = (\frac{2^1*7}{10^2}) + (\frac{2^2*7}{10^4}) + (\frac{2^3*7}{10^6}) + (\frac{2^4*7}{10^8})....$$
The k-th term of the above geometric progression appears to be $7*(\frac{2}{100})^k.$
This infinite geometric progression $7*\sum_{k=1}^\infty (\frac{2}{100})^k$ evaluates to $\frac17$, which confirms the observation.
Since this expansion was obtained through observation and not through a rigorous proof, it offers no insight as to why such a pattern exists. Any insights regarding this pattern will be appreciated. 
Edit: (rephrasing the question to make it more clear)
This infinite progression for $\frac17$ stems from a rather unusual pattern involving powers of $2$, and not from the recurring portion of the decimal ($0.142857$). The idea that the fraction $\frac17$ has anything to do with powers of two seems absurd, and I wanted to know if there is a justification for this pattern. Also, is this expansion involving powers of two unique to the fraction $\frac17$, or is there a more general way to apply this to any recurring decimal?
 A: Observe the pattern of $\dfrac1{49}=\dfrac2{100-2}$:
$$0.020408163265306122448979591836734693877551020408163265306\cdots$$
As a sum of terms $\dfrac{2^k}{100^k}$, it shows the successive powers of two in digit pairs, which is quickly destroyed by carries.
Now if you multiply by $7$, you will see the last two digits of $7\cdot2^k$, until carries hide them. But as the period is $6$ (which is maximal), only one pair is perturbed.

The phenomenon can be more striking with $\dfrac1{7142857}$,
$$0.00000014000000280000005600000112000002240000044800000896000\cdots$$
due to the fact that $7\cdot7142857=49999999$, so that you find the $8$ first digits of the multiples of $7$ by powers of $2$.
A: This is another solution that is much more simple:
Notice that $\frac{1}{7} = 0.142857 \cdots$ Since the series of digits $142857$ repeats every $6$ digits, then we can write $\frac{1}{7}$ as $142857*10^{-6} + 142857*10^{-12} + \cdots + 142857*10^{-6n}$. This can be written as a geometric series, with first term $a=142857*10^{-6}$, and common ratio $r = 10^{-6}$.
Using the geometric series formula $\frac{a}{1-r}$, we get $$\frac{142857*10^{-6}}{1-10^{-6}}.$$
We can multiply both top and bottom by $10^6$ to get $\frac{142857}{999999}$, which cancels out to $\frac{1}{7}$.
(another method is to write something like $10^6x = 142857.142857 \cdots (1)$, and $x = 0.142857 \cdots (2)$, and then subtract $(2)$ from $(1)$, but I think most people know this method already.)
A: In my opinion, the long division algorithm for computing the decimal expansion of $1/7$ gives a fairly clear reason why decimal expansions should repeat: at each step there are only finitely many possibilities for what the current remainder can be, so eventually you'll repeat a previous state, so everything you do after that point is identical to what you did the last time you were at that point.
(this is true even for terminating decimals; e.g. recall that $0.5$ is short for $0.5\overline{0}$)
