Show $y = \sum_{k=1}^\infty \frac{k^{k-1}}{k!} x^k$ satisfies $ye^{-y} = x$.

Hint: Substitute $u=-y$ then $-ue^{u}=x \implies ue^u=-x$, hence $u=W(-x)$, in which $W$ is the Lambert function. Use Lagranges inversion theorem to derive its Taylor series. Or simply look it up on the wiki page under asymptotic expansions. Note, that this equality is only true in the region of convergence.

Here is a combinatorial way to view this identity with generating functions.

The number of labelled unrooted trees with vertex set $[n]:=\{1, \ldots, n\}$ is $n^{n-2}$ (Cayley's formula). So the number of rooted trees is $\alpha_n := n\cdot n^{n-2} = n^{n-1}$. So $y = T(x) := \sum_{n=1}^{\infty} \frac{\alpha_n}{n!} x^n$ is the (exponential) generating function for the sequence $\alpha_n$, of the number of rooted trees.

The identity you want can be written $$ T(x) = x \cdot e^{T(x)} = x \left(1+ T(x) + T^2(x)/2!+T^3(x)/3!+\ldots \right). $$

So you want to show that the coefficient by $x^n/n!$ on the right hand side is also $\alpha_n$. Obviously, $\alpha_n = \sum_{k=0}^\infty \alpha_n(k)$, where $\alpha_n(k)$ is the number of rooted trees where the root has degree $k$ ($\alpha_n(k) = 0$ for $k \geq n$).

The term $x \cdot T^k(x)/k!$ is the generating function for the number of ordered pairs of $(v, F_k)$ where $v$ is a single vertex and $F_k$ is an unordered $k$-tuple of disjoint rooted trees. Now there is a natural bijection between such a pair and rooted trees (connect single vertex (the new root) with each root in $k$-tuple) where the root has degree $k$. In particular, the coefficient in front of $x^n/n!$ in $x \cdot T^k(x)/k!$ is the number of rooted trees on $n$ vertices where the root has degree $k$. That is, the coefficient is $\alpha_n(k)$.