I came across this proof while studying graph theory and although it instinctively makes sense, I have a problem with a part of it.
Proposition: A non-trivial simple graph $G$ must have at least one pair of vertices whose degrees are equal.
Proof: Let $G$ be a graph with $n$ vertices. Because $G$ is a simple graph, then by definition it doesn't contain multi-edges or loops. Therefore, there appears to be $n$ possible degree values for any vertex, namely $0, ..., n-1$. However, there cannot be both a vertex of degree 0 and a vertex of degree $n - 1$, since the presence of a vertex of degree 0 implies that each of the remaining $n - 1$ vertices is adjacent to at most $n-2$ other vertices. Hence, the $n$ vertices of $G$ can realize at most $n − 1$ possible values for their degrees [This is the part where I have troubles]. Thus, the pigeonhole principle implies that at least two of the $n$ vertices have equal degree.
Isn't this proof based on assumption that there exists a vertex with degree $0$. What happens if that is not case? Does the proof still hold? And why can the $n$ vertices of the graph realize at most $n-1$ possible values? I don't understand that step.