Let $\tau (G)$ be the number of spanning trees in graph $G$.
I will prove the following claims:
(1) if T is a tree, $\tau (T)=1$
(2) if G is derived from a tree T by replacing one edge of T by a multiple edge of multiplicity k, then $\tau (G)=k$
(3) $\tau (C_n)=n$
(1)Suppose a tree has two spanning trees. Consider two vertices x, y in T. Since there are two spanning trees, there are two distinct walks between x and y: $(x,p_1,...,p_n,y)$ and $(x,q_1,...,q_m,y)$. Then $(x,p_1,...,p_n,y,q_m,...,q_1,x)$ is a cycle in T. This is a contradiction because a tree by definition is a graph which contains no cycles.
(2) Let the edge $ab$ be in a tree be replaced by k edges. Since $ab$ is a bridge, so are the multiplied k edges. Hence, the graph obtained by choosing one of the k edges and deleting the k-1 edges is a tree. There are k such possible trees, and $\tau (T)=1$ so $\tau (G)=k$.
(3)$C_n$ has $n$ edges. Therefore, deleting one edge will result in a graph of $n$ vertices and $n-1$ edges, and this graph is a tree. There are n possible such tree, and $\tau (T)=1$ so $\tau (G)=n$.