computational geometry
We need to prove the following claim:
There exists a set of $n$ points such that a Voronoi cell of one of the points contains $n-1$ vertices.
I think in the following case the Voronoi cell for point $C$ contains $n-1$ vertices, but I'm unable to prove it.
$ P = \{ p_1, p_2,p_3,..,p_{n-1} \} \cup \{ C \} $
$S = \{p_i : 1 \leq i \leq n -1 \}$ are vertices of regular convex polygon and $C$ is its centre.
Take a supporting line $s$ of $\text{conv}(P)$ running through $p_i$. Then consider the ray $r$ emanating from $p_i$ that is orthogonal to $s$ and points away from $\text{conv}(P)$. Every point $x$ on this ray has $p_i$ as closest point of $P$. To see this consider the circle around $x$ that touches $p_i$, this circle is tangent to $s$ and therefore intersects $\text{conv}(P)$ only in $p_i$.
Since $r$ is unbounded and contained in $V(p_i)$, $V(p_i)$ is unbounded.
For (2), knowing $V(P)$ is not sufficient to determine $P$. For example, if $V(P) = \{5,15\}$ then $P$ could be $\{0,10,20\}$ or $\{1,9,21\}$ or $\{2,8,22\}$ etc.
Difficulty is that $V(P)$ gives you $n-1$ linear equations in $n$ unknowns, so system is underdetermined. Additional constraint that $p_{k+1} \ge p_k$ is not sufficient to give a unique solution.
Best Answer
I am assuming that you are talking about Voronoi cells on a plane with standard metric. I also guess that "cell contains $n-1$ vertices" means "the cell is a polygon which boundary has at least $n-1$ vertices".
Unfortunately theorem fails for $n = 2$ and $n = 3$ (unless you put a vertex in infinity). For $n \geq 4$ you are right. To prove that is suffices to show the following:
(1) is trivial for $n \geq 4$. To show (2) just consider a segment from $C$ to some of $p_i$ and some other $p_j$. Let $X$ be the midpoint of $Cp_i$, then by the triangle inequality and $|Cp_i| = |Cp_j|$ and $p_i \neq p_j$ we have that $|XC| = |Xp_i| < |Xp_j|$ so there exists a small ball where $C$ and $p_i$ are the only possible candidates for the closest vertex are $C$ and $p_i$ (just put a center at $X$ and take $r = \frac{\min_j|Xp_j|-|Xp_i|}{2}$). But $|XC| = |Xp_i|$ so cells of $C$ and $p_i$ share an edge of length greater or equal $2r$.
With (1) and (2) you can show that the cell of $C$ has at least $n-1$ edges and that it is bounded, i.e. each edge is a segment, which means that the boundary must have at least $n-1$ vertices and this concludes the proof.
Hope this helps ;-)