I tried to use induction to prove that the number of edges in a graph of $n$ nodes is $\frac{n (n-1)}{2}$. Does the following work?
Assume that a graph of $n$ nodes, where $n \geq 2$, has $\frac{n (n-1)}{2}$ edges. Now add a new node to the graph. By definition, in order for the graph to be complete again the new node needs to be connected to the $n$ previously existing nodes in the graph, hence an additional $n$ edges for a total of $\frac{n (n-1)}{2} + n = \frac{n^2 -n + 2n}{2} = \frac{n^2 + n}{2} = \frac{n (n+1)}{2} = \frac{(n+1)(n+1-1)}{2}$.
This shows that $P(n) \rightarrow P(n+1)$, where $P(n)$ is the statement that a graph of $n$ nodes has $\frac{n (n-1)}{2}$ edges.
Now we need to show that $P(n)$ is true for the base case of $n=2$. A graph of $2$ nodes has $\frac{n (n-1)}{2} = \frac{2 \cdot 1}{2} = 1$ edge, which is correct.
Yes, correct! I suppose you could make your base case $n=1$, and point out that a fully connected graph of 1 node has indeed $\frac{1(1-1)}{2}=0$ edges. That way, you can generalize your result to any $n \geq 1$.
In fact, the base even works for $n=0$ (no vertices means no edges, and indeed $\frac{0(0-1)}{2}=0$, and that doesn't change the inductive step (you have to always keep an eye on that as you move into the 'edge' cases, but you're ok here)