I would like to check if the following statement is correct (is there a canonical reference for this fact, if true?).
Take a tree graph $\mathcal{T}$ of size $N$, whose adjacency matrix is $A(\mathcal{T})$. [A tree is an undirected graph in which any two vertices are connected by exactly one path, or equivalently a connected acyclic undirected graph]. Consider the Laplacian matrix $$ L(\mathcal{T}) = D(\mathcal{T})-A(\mathcal{T}) $$ where $D(\mathcal{T})$ is the degree matrix of $\mathcal{T}$ [i.e., a diagonal matrix with vertex degrees on the diagonals], and let $\lambda_1,\ldots,\lambda_{N-1}$ be the $N-1$ nonzero eigenvalues of $L(\mathcal{T})$. Then $$ \prod_{i=1}^{N-1}\lambda_i = N\ . $$ [This would provide a polynomial-time, "easy" algebraic procedure to check whether a given adjacency matrix is the adjacency matrix of a tree - just compute its corresponding Laplacian and its nonzero eigenvalues].
My reasoning: by Kirchhoff's theorem, the number $\tau(\mathcal{G})$ of spanning trees of a graph $\mathcal{G}$ is $$ \tau(\mathcal{G})=\frac{1}{N}\prod_{i=1}^{N-1}\mu_i\ , $$ where $\mu_i$ are the nonzero eigenvalues of the Laplacian of $\mathcal{G}$. But if $\mathcal{G}$ is a tree, then it has only one spanning tree (the tree itself). Therefore, $\tau(\mathcal{T})=1$ and my claim would follow.
Is my reasoning correct?