Divisibility rule Example: $2^1$=2 --> $2\mid2$ 
If a number has their last digit divisible by 2, than the number is divisible by 2
$2^2$=4--> $4\mid2$,   $4\mid4$ 
If a number has their last two digit divisible by 4, than the number is divisible by 4
$2^3$=8--> $8\mid2$,   $8\mid4$ ,  $8\mid8$ 
If a number has their last three digit divisible by 8, than the number is divisible by 8
and so on...

How would you word this into a conjecture?

 A: Your conjuncture is essentially a true statement. The formulation of the statement is in @Ethan Bolker's answer. Here is also a prove for the statement:

Each number having $n+1$ digits can be written as follows:
$$10^na_n + 10^{n-1}a_{n-1} + \cdots+10a_1 + a_0 \tag 1$$
  
  
*
  
*Notice that we can extract $10$ from the first $n$ terms and the
  number will be $10m+a_0$ where $m = (10^{n-1}a_n + 10^{n-2}a_{n-1} +\cdots+a_1)$. Now, if $2$ divides this number, $2$ shall divide
  its form, $10m+a_0$. However $2|10$, hence, it will divide $10m$.
  Therefore $a_0$ must be divisible by $2$ so that the number is
  divisible by $2$.
  
*For the general case where $2^k$ divides $n$ if and only if $2^k$
  divides the last $k$ digits, extract $10^k$ from the first $n-k+1$
  terms, and follow the previous argument given that $2^k$ divides $10^k$ as $10^k = 2^k\times5^k$.

A: Your conjecture is (essentially) correct. You should be able to fill in the blank in

If a number has their last ?? digit divisible by ??, than the number
  is divisible by $2^k$.

But you are using the "$|$" symbol incorrectly. It means "divides", not "is divisible by", so
$$
4 \mid8
$$
but 
$$
8 \nmid 4
$$ .
Finally, the statements you write after the arrows are true when you turn them around, but they don't prove the conjecture. Were you supposed to do that?
A: Hint $\ \ 2^k\mid a\!+\!10^k b\iff 2^k\mid a\,\  $ by $\,\ 2^k\mid 10^k = 2^k 5^k$
Better  $\,\ a\bmod 2^k = (\underbrace{a\bmod 10^k}_{\large {\rm first}\ k\ {\rm digits}})\bmod{ 2^k},\ $ an example of the simpler multiple method.
Better $\,\ a\equiv b\pmod{\!10^k}\,\Rightarrow\, a\equiv b\pmod{\!2^k}$
Better $\,\ a\equiv b\pmod{\!mn}\,\Rightarrow\,  a\equiv b\pmod{\!n}\,\ $ by $\,\ n\mid mn\mid a-b$
e.g. $\bmod 1001\!:\ \color{#c00}{10^{\large 3}}\!\equiv -1 \,\Rightarrow\, a=12,013,002\equiv 12(\color{#c00}{10^{\large 3}})^{\large 2}\!+13(\color{#c00}{10^{\large 3}})+2\equiv 12\!-\!13\!+\!2\equiv 1$
so  $\ 7\!\cdot\!13=10^2\!-\!10\!+\!1\mid 10^3\!+\!1\,\Rightarrow\, a \bmod 13 = (a \bmod 1001)\bmod 13 = 1\bmod 13 = 1$.
In congruence language: $\ a\equiv 1\pmod{\!13j\!=\!\!10^3\!+\!1}\,\Rightarrow\, a\equiv 1\pmod{\!13}$
That's the idea behind one divisibility test for $13$.
