I’ve found this video very helpful in visualizing higher dimension solids (polytopes)
For super bonus extra fun, you can try using complex numbers instead of real numbers. Things are already pretty cray in 2 (complex!) dimensions [therefore something like 4 real dimensions] because a complex edge can intersect more than 2 vertices.
Now, instead of a tesseract, you have a configuration with 16 vertices and 8 4-edges. But there are many new possibilities…
Once discussion of the physical form of hyper cubes comes up I can’t help but cite the CM-1 and CM-2.
I remember in the 1980s IBM was experimenting with SIMD supercomputers with however many processors interconnected by a dynamically reconfigurable Clos network, so that at each machine cycle the connections can be set to a different permutation (so you can use a rectangular mesh of any dimension, hexagonal mesh, torus, whatever). The switches in this type of network are themselves recursively arranged in an interesting way:
I will not give the proof of this, but @senegoid’s post above provides the explanation. The number of k-dimensional “faces” (that’s the technical term) in an n-dimensional cube is the coefficient of x^ky^(n-k) in the binomial expansion of (x + 2y)^n. Since (x+2y)^5 = x^5 + 10x^4y + 40x^3y^2 + 80x^2y^3 + 80xy^5 + 32y^5, a 5-cube has 1 five dimensional face (itself), 10 4-dimensional faces (tessaracts) 40 3D faces (ordinary cubes), 80 2D faces (squares), 80 edges, and 32 vertices. Of course, the y is irrelevant; you could use (x + 2)^n instead.
At the risk of over-explaining this, replace (x+2y)^n above by (e+2v)^n. e stands for edge and v for vertex. An n-dimensional cube is, in a very precise sense, a product of n line segments and a line segment has one edge and 2 vertices. A k-dimensional face arises when k of the factors are edges and n-k of them are vertices and the coefficients of e^kv^(n-k) tells how many ways to do that.