Proof By Contradiction - Hamiltonian Paths and Cycles Was hoping if anyone had any way to prove the following claim using proof by contradiction

Let $G = (V, E)$ be a simple graph with at least one vertex, and let $G'$ be the graph formed by adding a new vertex $v$ and making it adjacent to every vertex in $V$.
Claim: $G$ has a Hamiltonian Path if and only if $G'$ has a Hamiltonian cycle.

I tried manipulating the definitions of each of the two (path vs. cycle), but didn't find much luck. Any thoughts?
Thanks!
Contradiction isn't the most natural way to do this, since there's a straightforward constructive proof. 

Suppose $G$ has a Hamiltonian path.  How can you extend that to a Hamiltonian cycle of&nbsp;$G'$?  Conversely, suppose that $G'$&nbsp;has a Hamiltonian cycle. Cover up $v$ with your finger&nbsp;&ndash; what do you see?

It's possible to rewrite the reasoning hinted at above as a proof by contradiction, but it's not very natural. ("Suppose $G$&nbsp;has a Hamiltonian path but $G'$&nbsp;has no Hamiltonian cycle. Oh, look, here's a Hamiltonian cycle in&nbsp;$G'$&nbsp;&ndash; contradiction.")