Let $G=(V,E)$ be a connected simple graph with vertex $v\in V$. Define $G'=(V',E')$ with $V'=V\smallsetminus{v}$ and $E'=\{e\in E|v\not\in e\}$. Prove that if $G$ is Hamiltonian then there exists a Hamiltonian path in $G'$.
Progress:
Assume that $G'$ doesn't have a Hamiltonian path. This implies that $G'$ isn't Hamiltonian either. There are a few cases:
1) $G'$ is not connected. Add $v$ and all the deleted edges back. In order for the new graph $(G)$ to be Hamiltonian there must exists a Hamiltonian cycle. However $v$ gets visited at least twice.
2) $G'$ contains a subdivision of $K_{3,3}$ or $K_5$ which implies that it'd be impossible to retrieve $G$.
We get a contradiction in both cases.
I'm not quite sure about this proof. It seems like a direct proof would be much more elegant.
When you delete $v$, a Hamiltonian cycle in $G$ becomes a Hamiltonian path in $G’$.