Suppose G is a connected graph and $v_0$$e_1$$v_n$ ... $e_n$$v_n$ is a circuit in G. Assume $v_{n-1}$ ≠ $v_n$. Prove that there is a simple circuit in G, $\tilde{v_0}$$\tilde{e_1}$$\tilde{v_1}$ ... $\tilde{e_m}$$\tilde{v_m}$ say, where $e_n$ = $\tilde{e_m}$.
I have no idea of where to approach this proof from. I would be grateful if someone could write it out for me and explain at the same time.
Induction on $n$.
If the given circuit is already simple, we are done. So assume it is not simple. Then there are $0\le i<j\le n$ with $v_i=v_j$ and $(i,j)\ne (0,n)$. If $j=n$, then we can take $(0,i)$ instead of $(i,j)$; in other words, we may assume that $j<n$. Then $v_0e_1v_1\ldots v_ie_{j+1}v_{j+1}\ldots e_nv_n$ is a circuit with the same last edge $e_n$, but shorter. By induction hypothesis, we may assume that there exists a simple circuit with last edge $e_n$.