Prove that split graphs are superfactorial I need to prove that split graphs family is superfactorial. 
Given a family of graphs $F$ we call $F_n$ to the number of graphs of $n$ labeled vertex in $F$
For example, in the family of complete graphs, given an integer $n>0$ we have that $F_n= 1$ for all $n$.
If $F$ is the family of simple paths we have that $F_3=3$, as the graphs induced by $\{(1,2),(2,3)\}$,$\{(1,2),(1,3)\}$ and $\{(1,3),(2,3)\}$ are different graphs.
I want to show that given $F = \{$the collection of split graphs$\}$ and a natural number $n$, there are at least $2^{\frac{n^2}{4}}$ split graphs of $n$ nodes, i.e. $F_n \geq 2^{\frac{n^2}{4}}$. 
How can I go about this?
 A: The number is exactly $\sum_{k=0}^n\binom{n}{k}(2^{k}-1)^{n-k}$ such graphs.
This is because there are exactly $\binom{n}{k}(2^{k}-1)^{n-k}$ graphs in which the clique has size $k$. Why? there are $\binom{n}{k}$ ways to select the clique. and then for each vertex $v$ outside the clique we must select a subset of vertices in the clique to which $v$ is adjacent (with the only restriction it is not the whole clique). So there are $2^k-1$ options and we must make this choice $n-k$ times.
So we have found the number of labeled split graphs on $\{1,2,3\ \dots n\}$. Letting $k=\lfloor n/2 \rfloor$ we get at least $\binom{n}{\lfloor n/2 \rfloor}(2^{\lfloor n/2 \rfloor}-1)^{\lceil n/2 \rceil}$.
Of course $2^a-1\leq 2^{a-1}$ and of course $\binom{n}{\lfloor n/2 \rfloor}\geq 1$.
So there is at least $2^{\lfloor n/2 \rfloor-1}2^{\lfloor n/2 \rfloor}$ split graphs. which is at least $2^{(\lfloor n/2 \rfloor-1)^2}$. For larg values of $n$ we have trivially that $\lfloor n/2 \rfloor-1\geq \frac{n}{3}$
So for large $n$ there is at least $2^{\frac{n^2}{9}}$ labeled split graphs on $\{1,2,3\dots n\}$. From here you can prove it is superfactorial:
$$2^{n(n/9)}=(2^{\sqrt n/9})^{n^{3/2}}>n^{cn}$$ 
for large enough $n$. Because $2^{\sqrt{n}/9}$ is eventually larger than $n$ (since $\sqrt n/9$ grows faster than $\log(n)$) and $n^{3/2}$ is eventually larger than $cn$.
