Spanning Trees of the Complete Graph minus an edge I am studying Problem 43, Chapter 10 from A Walk Through Combinatorics by Miklos Bona, which reads...

Let $A$ be the graph obtained from $K_{n}$ by deleting an edge. Find a formula for the number of spanning trees of $A$.

So how I approached this problem was by creating the Laplacian of A. I set the edge to be deleted as the edge between the first and second vertices in the graph. After an obscene amount of potentially dubious matrix operations, I received a result of...
$n^{n-3}*[2n^{3}-5n^{2}+3n \pm 1]$
Can anyone shed some light on this problem? I feel as I am approaching it the wrong way...
 A: 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.
A: Maybe an easier approach is to first count the number of spanning trees that containing that edge (which I think is equal to: $$\sum_{k=0}^{n-2} C(n-2,k) (k+1)^{k-1} (n-k-1)^{n-k-3}$$)
and subtract it from total $n^{n-2}$ spanning trees.
A: another approach:
Denote by $G_{i}$ the number of spanning trees on $K_n - e_i$ , where $e_i$ is the $i$-th edge in $K_n$ . Denote by $T_n$ the number of spanning trees on $K_n$ . We get the following equation wich I will explain bellow:
${\sum_{n=0}^{ n \choose 2} G_i} = ({n \choose 2} - (n-1))\cdot T_n$ 
Indeed any spanning tree of $K_n$ either contains the edge $e_i$ or it doesn't, so in the sum on the left hand side we actually count all spanning trees on $K_n$, but we are overcounting. When we evaluate $G_i$ we count all the spanning trees in $K_n$ evoiding $e_i$. Since every spanning tree has exactly $n-1$ edges, a spanning tree counted in the evaluation of $G_i$ (hence evoiding the edge $e_i$) is also evoiding ${n \choose 2} - (n-1)$ other edges of $K_n$, so we are counting it  ${n \choose 2} - (n-1)$ times in the sum on the left hand side. By a symmetric argument we conculde that this is the factor by wich we are overcounting the spanning trees on $K_n$. Finally, again by symmetry note that the $G_i$'s all have the same value (call it $G$). 
We get $ {n \choose 2}\cdot G = ({n \choose 2} - (n-1))\cdot T_n$. 
Simplification yields $ G = \frac{(n-2)}{n} \cdot T_n$ 
