Stuck at showing what the circumference of d-dimensional cube is So this is the exercise 2 of chapter 1 in Diestel's Graph Theory (5th ed.). Here is the description:
Let $d \in \mathbb{N}$ and $V := \{0, 1\}^d$; thus, V is the set of all 0-1 sequences of length $d$. The graph on $V$ in which two such sequences form an edge if and only if they differ in exactly one position is called the d-dimensional cube. Determine the average degree, number of edges, diameters, girth and circumference of this graph.
Below is whatever progress I have.
(Circumference) The circumference of the graph is $2^d$.
Proof: Induction on $d \geq 2$. Let $v_{i_1,\dots,i_d}, i \in \{0, 1\}$ be vertex $v$ that has the binary string $i_1,\dots,i_d$.
Case $d$ = 2: $G$ contains four nodes, $v_{0,0}, v_{0,1}, v_{1,0}, v_{1,1}$ for which the longest cycle is 4.
Suppose that he claim holds for $d = n - 1 \geq 2$.
Case $d = n \geq 2$.
(End).
What bugs me is that I think I understand the way the longest cycle can be formed, but I do not know how to argue about is mathematically. Essentially, we can think of the node $000\dots0$ and $111\dots1$ as the leftmost and rightmost vertex respectively (essentially two end vertices), and every other vertex in-between them. Then, each vertex has $d$ neighbours and there are $2^d - 2$ vertices between the two "end" vertices. Simply start from either "end" vertex, progress through the bottommost vertex to the other end and perform zigzagging back to the starting vertex. This monstrosity is my best attempt at drawing my idea. The red denotes how we form the cycle.
So how do I go about and finish this proof?
 A: Note that your title deals with the girth, while the body of your question is about the circumference; I’ll point you in the right direction to finish the argument in the body of your question.
One longest cycle on the $3$-cube is $000,100,110,010,011,111,101,001$. Notice that the second half of it can be obtained by reversing the first half and changing the last bit from $0$ to $1$. If we look just at the first two bits, the first half is a longest cycle on the $2$-cube: $00,10,11,01$, and the second half reverses that. This suggests that we might be able to use the same recipe to get a cycle of length $16$ on the $4$-cube from our $8$-cycle on the $3$-cube:
$$\begin{align*}
&0000,1000,1100,0100,0110,1110,1010,0010,\\
&0011,1011,1111,0111,0101,1101,1001,0001
\end{align*}$$
This idea is the basis for the induction step in a proof by induction that the circumference of the $n$-cube is $2^n$ for $n\ge 2$: for that step you should try to show that this procedure always produces a longest cycle in the $(n+1)$-cube if you start with a longest cycle in the $n$-cube.
