# Min. number of vertices in graph as function of $\kappa(G)$ and $diam(G)$

Question I found in "Introduction to Graph Theory" by Douglas B. West:

Let $G$ graph on $n$ vertices with connectivity $\kappa(G)=k \geq 1$. Prove that $n \geq k(diam(G)-1)+2$, when $diam(G)$ is the the maximal (edge) distance between a pair of vertices in $G$

Seems easy and I've seen the answer, too, which goes the same as thought: by taking the vertices the end vertices of the path of length $diam(G)$, and applying Menger's Theorem. I'm having a hard time understanding one part - it seems like all the paths need to be of the same length for the $n \geq k(diam(G)-1)+2$ to apply. Am I missing some part here?

Let's write $d$ for the diameter of $G$. Then there is some pair of vertices such that one path between them has exactly $d$ edges, and each path between them has at least $d$ edges. Menger says there are $k$ independent paths. So the number of vertices is at least $k(d-1)+2$.
I'm not sure why you think the paths all have to be the same length. They just have to have length at least $d$ - and, they do.
I don't think that all the paths have the same length, but that the maximal path length is $d$ (or am I misunderstanding the notion of $diam(G)$?). –  Pavel Feb 8 '12 at 19:25