Q:

Prove you can place binary words in a circle

A binary word of length $n$ is the series of $n$ elements that each one of them is $0$ or $1$.
Let $A$ be the set of all the binary words of length $n$.  
Prove you can place all the elements of $A$ in a circle such that two consecutive words have a difference of only ONE $1$ or ONE $0$ 
For example, $n=3$ :
 
I tried thinking of a recursive-like algorithm to do so but it did not work. I thought by starting at $000$ then $001$ however we cannot put $010$ because it is different from the previous by 2 digits number (0 changed to 1 and 1 changes to 0 from 001 -> 010)  so we move to $011$ and so on...
This is not a proof and it does not work either (the algorithm)   
I even tried to start proving by induction, but it sounds ridiculous because I can't think of how to prove that you can "draw a circle using n=k+1" using the assumption that you can draw it using words of length $n=k$ 
Edit something interesting I found:
if you draw $n=2$ then you get:  

Then add $0$ to the left of each word, you get:  

You get the beginning of the first image.. but I am not sure how to make it rigorous...   
I would appreciate your help! Thank you!

A:

Hint Prove the statement by induction.
Here is the inductive step $P(n) \Rightarrow P(n+1)$.
You know that you can place all the binary workds of length $n$ in a circle $x_1-x_2-x_3-...-x_{2^n}-x_1$.
Now, show that the following works
$$x_10-x_20-...-x_{2^n}0-x_{2^n}1-x_{2^n-1}1-...x_11-x_10$$
where $x_k0$ and $x_k1$ are the binary words made by $x_k$ followed by 0 or 1.
In other words: add a zero at the end of each word, go around the circle without closing it, and come back around the circle, this time adding a 1 at the end.

