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Could someone give a hint on this proof?

user300045
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Other than this answer:


Let $G$ be a graph with $n$ vertices. If $G$ is connected then as we know that a graph with $n$ vertices is a tree if and only if it is connected and has $n-1$ edges, it is not a tree, so it contains a cycle. If $G$ is not connected, one of its connected components has at least as many edges as vertices so this component is not a tree and must contain a cycle, hence $G$ contains a cycle.