  We use a sign-reversing involution to show that trees on the vertex set [n],
considered to be rooted at 1, in which no vertex has exactly one child are
counted by 1/n sum_{k=1}^{n} (-1)^(n-k) {n}-choose-{k} (n-1)!/(k-1)! k^(k-1).
This result corrects a persistent misprint in the Encyclopedia of Integer
Sequences.
