TITLE: Lower bound on # spanning trees in a connected graph
QUESTION [10 upvotes]: Are there are any good lower bounds for the number of spanning trees for a connected graph $G$ in terms of (for example) number of edges $E$ or number of vertices $V$ ? 
Are improved bounds available if one knows the number of bridges of the graph? (A bridge, aka a cut-edge, is an edge whose removal disconnects the graph)
EDIT:
Here is a simpler statement of the question. What is the minimum number of spanning trees possible for a graph with $m$ edges and $n$ vertices which is (1) 1-edge-connected or (2) 2-edge-connected?

REPLY [3 votes]: I don't now if this is trivial or not but given a connected graph $G=(V,E)$, there are are at least $|E|-|V|+2$ spanning trees.
Take any spanning tree T. There are $|E|-|V|+1$ edges $e$ that are not in this spanning tree. For any such $e$, $T+e$ has exactly one cycle. There exists an edge $f \in T$ s.t. by removing $f$ the cycle is broken and the tree is still connected. Each $T+e-f$ spanning tree is distinct as they include a different edge $e$.
I also would like to add something implicit to David's answer (but apparently I do not have reputation for commenting). The graph $G$ constructed that way includes the $K_{n-k}$ clique as a subgraph. By Cayley's Theorem, that clique has ${(n-k)}^{n-k}$ spanning trees and therefore  $G$ has at least ${(n-k)}^{n-k}$ spanning trees.