prove that k-cube has $2^k$ vertices, $k$ $2^k$ $^-$ $^1$ edges and is bipartite.

Question

The k-cube is the graph whose vertices are the ordered k-tuples of 0's and 1's, two vertices being joined if and only if they differ in exactly one coordinate.
Show that the k-cube has $2^k$ vertices, $k$$2^k$$^-$$^1$ edges and is bipartite.
Note : This question is from the book "Graph Theory With Applications" written by Bondy & Murty. Thanks in advance.

Answer 1

There are $2$ possibilities for each of the $k$ coordinates of a vertex, so there are $2^k$ vertices in total.

Each vertex has $k$ neighbours (why?) so the sum of the degrees of all the vertices is $k2^k$. But this counts each edge twice, so...

To show that the graph is bipartite, consider the coordinate sum of a vertex. If this sum is even, what can you say about the sums for neighbouring vertices? What if the sum is odd?