Number of edges Upper Bound

Given a simple graph with $n$ vertices and $m$ edges, then show: $m \le \binom{n}{2}$.

Obviously the equality holds when the graph is complete, and if you have less edges, then the inequality would intuitively hold. But how to show this rigorously?

An edge is uniquely determined by its set of endpoints, which is a subset of the set of vertices of size $2$, of which there are $\binom{n}{2}$. Hence this is an upper bound.

Let $V$ denote the set of vertices in the graph, and write $V = \{v_1, \ldots, v_n \}$.

$$m = \frac12 \sum_{v \in V} d(v) = \frac12\left( d(v_1) + \ldots + d(v_n) \right)$$

Let $v^* \in V$ such that: $d(v^*) \ge d(v), \forall \ v \in V$. Then:

$$m \le \frac12 \left( d(v^*) + \ldots + d(v^*) \right) = \frac{n}2 d(v^*)$$

But since the graph is simple with $n$ vertices, $v^*$ can't be linked to itself in a loop or linked to another vertex more than once. As a result, $d(v^*) \le n - 1$. Therefore:

$$m \le \frac{n(n-1)}2 = {n \choose 2}$$