Labelled unicyclic graph count ($C(n,n)=\frac{1}{2}(n-1)!{\sum }_{j=0}^{n-3}{n}^j/j!$)

A labelled unicyclic component has a unique cycle of length $k\ge 3$, with a rooted forest attached to its vertices. Choose its vertices in $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$ ways and its unoriented cycle in $(k-1)!/2$ ways. The labelled forest count with prescribed roots gives $k{n}^{n-k-1}$ attachments. Summing over $k$ and putting $j=n-k$ yields the displayed identity.