Min. number of vertices in graph as function of $\kappa(G)$ and $\operatorname{diam}(G)$ 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 $$n \geq k(\operatorname{diam}(G)-1)+2$$ where $\operatorname{diam}(G)$ 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 $\operatorname{diam}(G)$, 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 $n \geq k(\operatorname{diam}(G)-1)+2$ 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. 
