A sign-reversing involution to count labeled lone-child-avoiding trees
  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. 
David Callan
math.CO