I am studying graph theory and my text book gives a proof for an n-cube graph being connected that I find really weird/confusing. Is there a simpler proof to show that the n-cube is connected.

The n-cube graph is the graph where the vertices {0,1} strings of length n. Two strings are adjacent if they differ is exactly one position. 

For example the 2-cube graph has vertices V:{00, 01, 10, 11} and edges: 
(00 <-> 01),
(00 <-> 10),
(01 <-> 11),
(10 <-> 11)

This is simply a square. What kind of proof can I use show that it is connected? I am currently thinking induction.

**My rough try at a Proof**

Proof By **Induction**

Base Case ( n=1 ) 
Then graph has two vertices, 0 & 1 which are connected (since they differ by one). Assume than P(n) is true. Prove P(n+1). Lets remove the last digit. From the inductive hypothesis we know that P(n) is connected. Add the last digit back in. We can partition P(n+1) into $A\_0$ which contains all the vertices with a 0 as the last digit; $A\_1$ which contains all the vertices with 1 as the last digit. Since $A\_0$ and $A\_1$ are essentially P(n), we know they are connected from the inductive hypothesis. Since $A\_0$ and $A\_1$ are connected we can conclude that P(n+1) is connected! 

**Is this correct?**