Prove that a simple graph $G$ with $n$ vertices, $n \ge 2$; is complete if and only if $\kappa(G)[\text{vertex connctivity}]=n - 1$:
proof. n=1. It is trivially true. Suppose we assume result hols for $n=k$, We need to prove this result hold for $n=k+1$. Consider $K_{k+1}$ complete graph, removal of one vertex resuts $K_k$. We know that the $k-1$ vertex required to remove the graph disconnected. So, $\kappa(K_{k-1})=k-1+1=k$. Hence, by induction, result holds for every $n\in \mathbb N$. Am I correct?
For the statement $\kappa(G) = n-1$ implies $G=K_n$, again use induction.
The base case is simple.
The induction hypothesis is then that if $G$ has $k-1$ vertices and $\kappa(G) = k -2$, then $G = K_{k-1}$.
Now, take a graph $G$ with vertex connectivity $\kappa(G) = k-1$. Pick an arbitrary vertex $v$, now $\kappa(G-v) = k-2$ and has $k-1$ vertices, so that $G-v$ is the complete graph on $k-1$ vertices by the induction hypothesis. We can see that $\operatorname{deg}(v) = k-1$, because otherwise, the deletion of less that $k-1$ vertices from $G$ would disconnect $v$ from $G$, contradicting the fact that $\kappa(G) = k-1$. Therefore, $G-v$ is $K_{k-1}$ and $v$ is connected to every vertex in it. Thus $G = K_k$.