Proof of the number of edges of a complete graph

Proof

1. The complete graph $K_n$ has vertex set $V = \set{1, \dots, n}$, and any two distinct vertices are adjacent.
2. Each edge of a graph is a $2$-element subset of $V$. By step 1, every $2$-element subset $\set{u, v}$ of $V$ is an edge of $K_n$.
3. So the edges of $K_n$ are exactly the $2$-element subsets of $V$. By the number of k-element subsets, there are $\binom{n}{2}$ of them.
4. If $n \ge 2$, the factorial formula gives $\binom{n}{2} = \frac{n!}{2!\,(n-2)!} = \frac{n(n-1)}{2}$.
5. Let $n \in \set{0, 1}$. The recursion that defines the binomial coefficient gives $\binom{0}{1} = \binom{0}{2} = 0$. Hence $\binom{1}{2} = \binom{0}{1} + \binom{0}{2} = 0$.
6. For $n \in \set{0, 1}$ also $\frac{n(n-1)}{2} = 0$. So in all cases $K_n$ has $\binom{n}{2} = \frac{n(n-1)}{2}$ edges.