Spanning Trees of the Complete Graph minus an edge

There's no need to consider the Laplacian. We can obtain this by a simple symmetry argument.

Every edge of the complete graph is contained in a certain number of spanning trees. By symmetry, this number is the same for each edge, call it $k$. Let us now count the total number of edges in all spanning trees in two different ways.

First, we know there are $n^{n-2}$ spanning trees, each with $n-1$ edges. Therefore there are a total of $(n-1)n^{n-2}$ edges contained in the trees.

On the other hand, there are $\binom{n}{2} = \frac{n(n-1)}{2}$ edges in the complete graph, and each edge is contained in precisely $k$ trees. This means there are a total of $\binom{n}{2}k$ edges.

This gives us $$(n-1)n^{n-2} = \binom{n}{2}k$$ which upon simplification gives $k=2n^{n-3}$.

If we delete an edge, then we effectively remove the set of all spanning trees containing that edge. By assumption that number is $k$. Therefore there will remain $$n^{n-2} - k = n^{n-2} - 2n^{n-3} = n^{n-3}(n-2)$$ total spanning trees.