Is it always possible to orient an $n$-dimensional object (say an $n$-cube) so that, when projected onto a $(n-1)$-dimensional flat surface, no two of its vertices intersect with one another in the projection, for $n \geq 2$.
If this is true, applying it $n-1$ times, we should be able to represent the $n$-dimensional object as (in the case of the $n$-cube, $2^n$) points on a line. It seems like it should be true, and perhaps trivially so, but how would one go about proving it?
I've tried thinking about concepts related to linear algebra, like that one has one more basis vector to play with in the higher dimension, but the exact argument eludes me.
Here's the case of $n=2$:
Let as order pairs of vertices to $N={n\choose 2}$ pairs $(X_i,Y_i)$. If $p(X_0)=p(Y_0)$, you choose any rotation that separates them. Inductively, if all pairs up to the $i$th one are separated and $X_{i+1},Y_{i+1}$ is the $(i+1)$th pair, then you choose a rotation so that $p(X_j)\neq p(Y_j)$ for all $j\leq i+1$. This is possible, as you can separate $X_{i+1}$ and $Y_{i+1}$ by arbitrary small rotations (small=close to identity) and the space of rotations for which $p(X_i)\neq p(Y_i)$ is open.