i was trying to proof this upper bound in the number of spanning trees $t(G)$ of an r-regular graph G (and discuss what happens with equality)
$t(G)$ $\leq$ $\frac{1}{n}$$(\frac{rn}{n-1})^{n-1}$
What i tought was to use the matrix-tree theorem. What we know is that
$t(G)$=$\frac{1}{n}$$\prod_{i=1}{\lambda_i}$ where $\lambda_i$ is an eigenvalue of the laplacian matrix of the graph(that is defined as the diagonale degree matrix minus the adjacency matrix of the graph). Now i think to prove that the product of than eigenvalues is less that power to n-1 but i don't know how to continue, i have to use the fact that is r-regular i think. Plus i googled around to proof this and i have found this document: https://cs.anu.edu.au/people/Brendan.McKay/papers/SpanningTrees1983.pdf Where in theorem 1.1 gives you three reference that i am not able to find anywhere online. Except for this reference book "Algebraic graph theory by Gibbs" that is a little bit expensive for just one proof. Any ieas?
You can simply use the inequality of the arithmetic-geometric mean which in your context states that $$ (\prod_{i=2}^n \lambda_i)^{n-1} \leq \frac{1}{n-1} \sum_{i=2}^n \lambda_i.$$
Now the sum of the eigenvalues is the trace of the respective matrix which in your case is twice the number of edges. For a $r$-regular graph that is $rn$ as needed.