Number of Nilpotent matrices. I am trying to find number of nilpotent matrices over the field of reals with entries as $0$ and $1$ only. I tried it for order $2$ its comes only $2$ as diagonal elements are always $0$. Is there general formula to find number of such types of matrices? Thanks in advance. 
 A: Some numerical results. For $n=1$ there is $1$ solution, for $n=2$ there are $3$ solutions, for $n=3$, $25$ sol, for $n=4$, $543$ sol, for $n=5$, $29281$ sol.
It is the beginning of the sequence $A003024$ that gives the number of $n\times n$ real (0,1)-matrices with all eigenvalues positive. 
cf. http://oeis.org/search?q=1%2C3%2C25%2C543%2C29281&sort=&language=english&go=Search
EDIT 1. $A003024$ is given by the recurrence formula: 
$a_0=1,a_n=\sum_{k=1}^n(-1)^{k-1}(_{k}^n)2^{k(n-k)}a_{n-k}.$
EDIT 2. The sequence $(a_n)$ is also the number of DAGs (directed acyclic graphs) on $n$ labeled nodes. Because a DAG cannot have self-loops, its adjacency matrix $A$ must have a zero diagonal. In the same way, a DAG has no loops of length $k\leq n$, that implies that the diagonal of $A^k$ is $0$. Finally, an adjacency matrix of a DAG is a $(0-1)$-matrix $A$ that is nilpotent; conversely, $A$ is a nilpotent $(0-1)$-matrix iff the diagonals of the $(A^k)$ are $0$. 
A: Hint:
The only eigenvalue of a nilpotent matrix is $0$, so, by Jordan canonical decomposition, any nilpotent matrix is similar to a matrix that  is a block diagonal matrix
$$
\begin{bmatrix}
A_1&0&\cdots&0\\
0&A_2&\cdots&0\\
\cdots\\
0&0&\cdots& A_k
\end{bmatrix}
$$
with blocks that are ''shift'' matrices of the form 
$$
A_i=
\begin{bmatrix}
0&1&0&\cdots&0\\
0&0&1&\cdots&0\\
\cdots\\
0&0&0&\cdots&0
\end{bmatrix}
$$
