Is there a formula for the number of labeled forests with $k$ components on $n$ vertices?

A formula as a single sum is $$f_{n,k} = \binom nk \sum_{i=0}^k \left(-\frac12\right)^i (k+i)\,i!\, \binom{k}{i}\binom{n-k}{i} n^{n-k-i-1}.$$ This formula can be found in J. W. Moon's Counting Labelled Trees, Theorem 4.1. He attributes it to A. Rényi, Some remarks on the theory of trees, Publications of the Mathematical Institute of the Hungarian Academy of Sciences 4 (1959), 73--85.

To prove it, let $T= \sum_{n=1}^\infty n^{n-1} x^n/n!$ be the exponential generating function for rooted trees and let $U=\sum_{n=1}^\infty n^{n-2} x^n/n!$ be the exponential generating function for unrooted trees. It is well known that $$T^k = k! \sum_{n=k}^\infty \binom nk k n^{n-k-1}\frac{x^n}{n!}.\tag{$*$}$$ Then $U = T-T^2/2$ so the exponential generating for forests of $k$ trees is $$\sum_{n=k}^\infty f_{n,k} \frac{x^n}{n!} = \frac{U^n}{n!}=\frac{(T-T^2/2)^k}{k!},$$ and the formula for $f_{n,k}$ follows by expanding the right side by the binomial theorem and using $(*)$.

A straightforward formula: $f_{n,k}$ is the coefficient of $x^n$ in $$\frac{n!}{k!}\cdot\left( \sum_{i=1}^{\infty} \frac{i^{i-2}x^i}{i!}\right)^k$$ (clearly, the sum here can be restricted to first $n$ terms).