This is motivated by the definition of simple polytopes: if all vertices of a $d$-dimensional convex polytope $P$ are in exactly $d$ edges (i.e. $1$-dimensional faces of $P$), then $P$ is simple.
I struggle to show that no vertex of $P$ can be in fewer than $d$ edges. Clearly, if this would be the case for some vertex $v$, the (translated) cone generated by $v$ and its adjacent vertices would have a lower dimension than $d$.
So if I can only show that the cone contains $P$, I would be done. That appears obvious but I've spent some time struggling with no luck. I feel I am missing something obvious.
Edited to add: I use "polytope" to mean the convex hull of finitely many points in some Euclidean space. And, for the sake of completeness, when I speak of $P$ being $d$-dimensional, I mean that its affine hull is $d$-dimensional, not the ambient space.