Does the $K_n \star K_n\backslash\{e\}$ contain $K_{k\geq n}$? 
Q1: Does the $K_n \star K_n\backslash\{e\}\star K_n\backslash\{e\}\star\cdots\star K_n\backslash\{e\}$ contain at least a $K_{k\geq n+1}$?
Q2: Is the every $K_{n+1}$-free as above?

My definition of $A\star B$ is the following:
Draw all edges between $A$ and $B$ such that each vertex of $B$ adjacent to $n-1$ vertices of $A$ and degree of every vertices of $A$ arising from $B$ be equal. for example If $A,B$ be $K_3$ then $A\star B$ is as follow:

As shown is this picture, degree of of every vertices of $A$ arising from $B$ are equal $2$ (green lines)
 A: It's possible even for $K_n \star K_n \star \dots \star K_n$ to avoid containing a clique of size $n+1$. (And any extra edges removed don't help.)
Give each vertex in $K_n$ its own label between $1$ and $n$. The vertices of $\underbrace{K_n \star K_n \star \dots \star K_n}_k$ are the vertices of $k$ copies of $K_n$, so they each inherit this labeling (there will be $k$ vertices labeled with $1$, $k$ vertices labeled with $2$, and so on). So let our rule for choosing how to construct the next iteration $$K_n \star (\underbrace{K_n \star K_n \star \dots \star K_n}_k)$$ be as follows: connect a vertex of the first $K_n$ with label $i$ to every vertex of the $k$-fold $\star$-sum that does not have label $i$.
In the resulting graph, no two vertices with label $i$ are connected, and so a clique can include at most one vertex with label $i$. Therefore no clique has size more than $n$. (There's lots of cliques of size $n$, of course.)
In fact, by Turán's theorem this graph has the maximum number of edges possible for a graph with no clique of size $n+1$. This tells us that whenever $K_n \star K_n \star \dots \star K_n$ is $K_{n+1}$-free, it must be a Turán graph on the appropriate number of vertices: a complete $n$-partite graph $K_{k,k,k,\dots,k}$. We can label each vertex with the part that it's in, and obtain the labeling above. So, up to isomorphism, the construction above is unique.
The vast majority of ways to take the $\star$-sum at each iteration don't follow this rule, and will produce large cliques. For an extreme example, if the rule is instead "connect a vertex of the first $K_n$ with label $i$ to every vertex of the $k$-fold sum that does not have label $i + n/2$ modulo $n$" then take every vertex with label $1, 2, \dots, n/2$, and you get a clique of size $kn/2$ in the $k$-fold sum.
If we replace $K_n$ by $K_n \setminus \{e\}$ at every step, not much changes: in the $k$-fold sum, this only loses us $k$ of the edges, so it can reduces the size of any given clique by at most $k$. Going from $kn/2$ to $kn/2-k$ doesn't make much of a difference in the long run. And generally we will be able to remove an edge that does not hurt the largest clique at all, if we want to.
