Applying transitivity to digraphs question. QUESTION:
Consider all possible forms of a digraph with 2,3,4...n vertices.
Now let f(A) be the number of edges in A. 
Also, let A^2 be the transitive of A, A^3 the transitive of A twice, A^4 the transitive of A three times, and so on.
Prove that f(A), f(A^2), f(A^3),.... f(A^n) converges to a single number, ie doesn't oscillate. 
ATTEMPT: 
I tried to prove it by cases with 2 vertices so far (16 digraphs) and it holds true that f(A), f(A^2), f(A^3),.... f(A^n) convergers to a single number at each case. With just 3 vertices, there are hundreds of cases so proving this concept onward by cases isn't time-efficient. That being said, I don't know how to move forward so any tips or advice would be really appreciated.
Note:
I am still fairly new to upper-level math so I apologize in advance for any way undesirable way I format or tag this post. Thank you for reading! <3 
 A: Unluckily, I'm not coming up with anything at the moment. Here are a few properties, maybe they help:
Let $ G= (V,E) $ be a graph. Let's define the transitivity-operation as 
$f((V,E)) := (V,E')$, with $E':= \{(x,y)\in V^2\mid \exists a\in V: (x,a),(a,y)\in E \}$
The key property is that if 
$$x_1\to x_2\to ...\to x_n$$
 is a path in $G$, then 
$$x_1\to x_3\to ...\to x_{n-1}$$
 is a path in $f(G)$.
Let's call $x\to y$ path connected, if there exists some path $x\to y$.


*

*If at some iteration a path connection $x\to y$ is lost, then it is impossible to restore it.

*The path connection only is maintained into the next iteration if there is a path of even length from $x$ to $y$.

*For a path connection $x\to y$ to never be lost, one needs to be able to construct an arbitrarily long path from $x\to y$ (in $E$) with even length. This is possible exactly if there's a path $x\to y$ that passes through a circle with an odd number of vertices.

*An edge $x\to y$ will exist permanently after some iteration exactly if we can create a path of length $2n$ for all $n\ge k$ for some $k$.


Partial result: Any graph $G$ without odd cycle has $$\lim_{n\to\infty} M( f^n(G)) =  0$$
Partial result:
Let $P$ be the adjacency matrix of the graph. Then we have $$(P^n)_{i,j} >0 \Leftrightarrow \text{There's a path from $i$ to $j$ with length } n$$
Especially this means that $M(f(G^n))$ converges exactly if the number of zero-fields in $P^n$ is at some point constant.
