Can a regular tetrahedron cast a square shadow given parallel light rays? How do you prove this? And, more generally, can a $n$-simplex cast a hypercube shadow in one lower dimension?
 A: Answer is "yes". Take two non-adjacent edges of the tetrahedron and construct a plane, parallel to both of them. This will be your projection plane. The projection is supposed to be orthographic - all projection lines are perpendicular to the projection plane.
The tetrahedron will be viewed from above as (the invisible edge is drawn by dotted line):

Informal proof: these two edges are perpendicular, have the same length and intersect in the middle - so they are diagonals of a square.
As for your general question about $n$-simplex - I appreciate the @IvanNeretin answer, given in comments. The answer is "no" for all $n \gt 3$, because the $n$-simplex has $n+1$ vertices, so any its projection will have this number of vertices, or less. However the number of vertices of $(n-1)$-dimensional hypercube is $2^{n-1}$, which is more than $n+1$ for $n \gt 3$. 
A: Yes.
Viewing direction is along center of two skewed sides onto a plane perpendicular to it in orthographic projection. It also passes through tetrahedron center. The viewing point is at infinity.
Spherical symmetry restraint of four connected lines of equal projected length makes for a square for such a viewing direction. Mathematica projection is slightly off due to finite viewing distance and due to angle made to such a direction.

