Proving that in a simple graph with $n$ vertices, two vertices have the same degree I want to prove this claim using contradiction:

Let $G$ be a simple graph with $n$ vertices. Prove by contradiction that $G$ has two vertices having the same degree (Consider the vertex of smallest degree and vertex of largest degree). 

I think I know where to start. The vertex of the largest degree is $n-1$ in this graph and the smallest is 1. How can I use contradiction to prove the statement above? I studied the proof of this using pigeonhole but I wanna use contradiction this time.
 A: Suppose such a graph exists. Each vertex in the graph can have a degree from $0$ to $n-1$ (simple graphs do not forbid a degree-$0$ vertex, connected graphs do). Since this range spans $n$ values in total and each vertex degree is different, the degrees are distributed one-per-vertex. In particular, there must exist a vertex with degree $n-1$ and one with degree $0$.
Now the degree-$n-1$ vertex is connected to all other vertices because the graph is simple, including the degree-$0$ vertex. But the latter is not connected to any vertex, which is a contradiction. Therefore two vertices in the graph must have the same degree.
A: Take a graph $G$ where all vertices have distinct degrees. So, $$d(1)<d(2)<\cdots<d(n),$$
and $$d(1)>0, d(2)>d(1)>0, \ldots, d(n)>d(n-1)>\ldots>0.$$
Hence we have just one vertex of degree $\geq n-1$.
However, in a simple graph with $n$ vertices a vertex can have a maximum degree of $n-1$. So, $d(n)=d(n-1)$. So, our assumption of having all degrees different was wrong and we have a contradiction. 
So, in a simple graph $G$, we must have at least $2$ vertices of the same degree.
