# Number of spanning trees in undirected simple graph

What is the number of spanning trees in an undirected simple graph?

My attempt:

Let $$m$$ be the number of edges in a simple graph, and let $$n$$ be the number of vertices.

Then number of spanning trees is $$\binom{m}{n-1}$$ minus the number of cycles of length $$n-1$$.

I read on Wikipedia that the number of spanning trees in the complete graph $$K_n$$ is $$n^{n-2}$$.

According to the formula I stated above, it should be $$\binom{n(n-1)/2}{n-1} - \binom{n}{n-1} (n-1)!$$.

How do I show that this is equal to $$n^{n-2}$$ for $$K_n$$?

You cannot show this since it isn't true. I encourage you to try out some actual numbers (e.g. $$n=2$$) and see for yourself that the numbers don't match. The problem is that what you should be subtracting is not the number of cycles of length $$n-1$$, but rather the number of collections of $$n-1$$ edges which contains at least one cycle. Such a collection need not be a cycle of length $$n-1$$.
In addition, the formula for the number of cycles of length $$n-1$$ is wrong. The number is $$n(n-2)!$$ rather than your $$n(n-1)!$$, and even this formula is only valid for $$n \geq 4$$ (for smaller $$n$$ there are no such cycles). However, even after replacing $$n(n-1)!$$ by $$n(n-2)!$$, the numbers don't match.