Number of directed graphs without isolated vertices I need to count the number of simple directed graphs, with n vertices, without isolated vertices. There is additional note in task saying that we assume that two graphs are different if there are two vertices which are connected in first but disconnected in the second.
This whole 'no isolated vertices' thing connected with the fact that we are talking about directed graphs makes it way more complicated than different cases I've found on the Internet.
Basically the only tip I've got from my teacher is to use inclusion–exclusion principle to eliminate isolation cases but I don't really know how to do that.
 A: For the case  of $n$ labeled nodes  we find using PIE  the closed form
$$D_n= \sum_{p=0}^n {n\choose p}  (-1)^p 4^{n-p\choose 2}.$$ This will
produce $$0, 3, 54, 3861, 1028700, 1067510583, 4390552197234, \ldots$$
which  points  to  OEIS A054545  where  it
appears we have a match. 
For the unlabeled case observe that the number $F_n$ of non-isomorphic
digraphs      was     computed      at     the      following     MSE
link.    We   then
obtain for the number of digraphs with no isolated nodes
$$D_n = F_n - F_{n-1}$$
which will produce
$$0, 2, 13, 202, 9390, 1531336, 880492496, 1792477159408,\ldots$$
again       a       match,       this      time       with       OEIS
A053598. 
More data concerning PIE. The nodes  of the poset for use with PIE
correspond to  all subsets  $P$ of  vertices of  the $n$  vertices and
represent labeled  digraphs where the  vertices in $P$,  plus possibly
some  other vertices,  are isolated.  The weight  on the  the digraphs
specified  by $P$  is $(-1)^{|P|}.$  This means  the digraphs  with no
isolated vertices have weight one because they appear only in the node
that corresponds to $P=\emptyset.$ Digraphs with a set $Q$ of isolated
vertices where  $|Q|\ge 1$ appear in  all nodes $P\subseteq Q,$  for a
total weight of
$$\sum_{P\subseteq Q} (-1)^{|P|} =
\sum_{p=0}^{|Q|} {|Q|\choose p} (-1)^p = 0,$$
i.e. zero.  The  cardinality of the set of  diagraphs corresponding to
node $P$  is $4^{n-|P|\choose  2}.$ We  now sum  the weights  over all
digraphs,  collecting  the contributions  from  all  nodes where  they
appear.  We  already know that  this will  assign weight one  to those
with no  isolated vertices and  zero otherwise, providing  the desired
count. There  are ${n\choose p}$ sets  $P$ of $p=|P|$ nodes  and there
are $4^{n-p\choose 2}$  digraphs at these nodes  with weight $(-1)^p$,
concluding  the derivation  of the  formula using  PIE. Note  that the
count is the same regardless of  whether we iterate over all digraphs,
collecting the  weights from the  nodes or over all  nodes, collecting
the weights of all digraphs.
