Does this identity with binomial coefficients have a combinatorial interpretation? I found the identity for $n\ge 1$:
$$
(n-1)n^{n-1} = \sum_{j=1}^{n-1} \binom{n}{j} j^j (n-j)^{n-j-1}
$$
or equivalently, if we say $0^0 = 1$, we have
$$
n^n = \sum_{j=0}^{n-1} \binom{n}{j} j^j (n-j)^{n-j-1}
$$
This identity is easy to show using generating functions (as I couldn't find this identity anywhere online, I include this proof below). However, it looks suspiciously like something that would have a combinatorial interpretation, since $n^n$ is the number of functions from a set of $n$ elements to itself, and the binomial coefficients could be taken to represent numbers of subsets of the same set. My question is, does this identity have a combinatorial proof or some meaningful combinatorial interpretation at least?
Proof using generating functions: The power series for the principal branch $W(x)$, the Lambert W function, is
$$
W(x) = \sum_{n=1}^\infty \frac1{n!} (-n)^{n-1} x^n
$$
and $W'(x) = \frac{W(x)}{x(1+W(x))}$. Thus we have:
\begin{eqnarray}
\sum_{n=1}^\infty \frac1{n!} n^{n-1} x^n &=& -W(-x) \\
\sum_{n=1}^\infty \frac1{n!} n^n x^n &=& x{\frac{d}{dx}}(-W(-x)) = \frac{-W(-x)}{1+W(-x)}\\
\sum_{n=0}^\infty \frac1{n!} n^n x^n &=& 1 + \frac{-W(-x)}{1+W(-x)} = \frac{1}{1+W(-x)}
\end{eqnarray}
We therefore have the product 
$$
-W(-x) \cdot \left(\frac{1}{1+W(-x)}\right) = \frac{-W(-x)}{1+W(-x)}
$$
which in series form looks like
$$
\left(\sum_{n=1}^\infty \frac1{n!} n^{n-1} x^n\right)\left(\sum_{n=0}^\infty \frac1{n!} n^n x^n\right) = \sum_{n=1}^\infty \frac{1}{n!}\left(\sum_{j=0}^{n-1} \binom{n}{j} j^j (n-j)^{n-j-1}\right) x^n = \sum_{n=1}^{\infty} \frac1{n!} n^n x^n
$$
 A: Recall the classic Cayley's Result that there are $n^{n-2}$ labelled tree on $n$ vertices. https://en.wikipedia.org/wiki/Cayley%27s_formula
Choose one of the vertices of a labelled tree and call it the (first) root ($r_1$); it is clear that there are $n^{n-1}$ rooted trees.
Next choose a second vertex ($r_2$) (possibly the same as the first) and call it the second root; It is clear that there are $n^n$ such objects and we will refer to these as bi-rooted trees. Thus we have a combinatorial interpretation of the LHS.
There is a unique path from the first root to the second root; lets denote this sequence of vertices by $r_1 x \cdots r_2$.
If $r_1=r_2$ the argument below becomes the $j=0$ case.
Now remove the edge between $r_1$ and $x$; this will break the tree into two smaller trees. 
Let $j$ be the number of vertices in the tree that contains the $r_2$ vertex; along with the $x$ vertex this is a bi-rooted tree of which there are $j^j$. The second tree will contain $n-j$ vertices and is a rooted tree (of which there are $(n-j)^{n-j-1}$). Finally there are of course $n$ choose $j$ ways to distribute the labels and so we have
\begin{eqnarray*}
n^n = \sum_{j=0}^{n-1}\binom{n}{j} j^j (n-j)^{n-j-1}.
\end{eqnarray*}
