Show that if $G$ is a connected graph of order $n≥2$ and $k$ is an integer with $1≤k≤n-1$, then $G^k$ is $k$-connected.
I'm not sure if I'm on the right track, but I tried to prove this by induction
Base: $n=2$ then $k=1$ and we have $G^1=P_2$ clearly $1$-connected
Inductive: Assume this is true for $n=l$ then $G^k$ is $k$-connected with $1\le k\le l-1$, I'm not sure this give me any useful information to to show this is true for $n=l+1$
Don't use induction.
Suppose $S$ is a vertex cut in $G^k$ with less than $k$ vertices and $v$ and $w$ are two vertices separated by $S$.
If $d_G(v,w)\leq k$, then $v$ and $w$ are neighbours in $G^k$, so they cannot be separated by $S$.
Otherwise the shortest path $P$ (in $G$) has more then $k$ internal vertices. In order to avoid that $P-S$ still contains a path from $v$ to $w$ (in $G^k$) $S$ must take away at least $k$ consecutive vertices and this is impossible.
Therefore $G^k$ must be $k$-connected.