Can anyone clearly prove Euler’s formula?

Answer

I surfed once found a website with 20 proofs of the formula
http://www.ics.uci.edu/~eppstein/junkyard/euler/
Some proofs require quite a bit of math, others less but are usually longer.
I like the proof number 6 the best

But here’s an argument that requires no additional math, but is only valid for the simplest cases. We assume that the body is singularly connected. It therefore consists of 1 piece, and contains no perforations. A torus is therefore not sufficient.
The idea is that we do some “operations” on a body that don’t affect the validity of the formula. So you start from a field L1 and you show that if the formula holds for L1, then the formula also holds for the body L2, the result of the operation and vice versa. Thus we are going to reduce a field L to an ordinary tetrahedron, for which the formula is clearly correct. And because the operations done preserve the validity, the formula is also valid for the original body.

What can we do with the body you start from?

First, any side that has more than three vertices is divided into triangles by connecting existing angles of the side. You can also use this to divide any concave sides into triangles. What is the consequence?
If you connect a connection between two existing angles in a side you have added 1 edge, but the number of sides also increases by 1, because you have divided an existing side into two pieces.
Now take the formula to be proved
Z + H = R + 2
If the formula is correct in our starting body, it will also remain correct if you add such an extra rib because 1 is added to the left and right in the formula. Conversely, removing such a side again retains the correctness because 1 rib disappears on the right and 1 side face on the left because you are actually merging two of them.
=> our body now consists of only triangles

Second. Now take an edge, and move one vertex to the other. We then lose:
– one vertex (because you put two in the same place)
– three ribs (see figure: from 5 to 2)
– two side faces, namely the two triangles adjacent to the disappeared edge

Now take again the formula to be proved
Z + H = R + 2
This action removes 3 on the left (2 sides + 1 vertex) and on the right also (3 ribs)
So if the formula is correct for one body, it is also correct for the other body. After all, it also applies in the other direction, by opening the vertices that we just merged again

Finally: remove enough vertex until you have 4 left. Then you have a tetrahedron, of which you know the formula holds.
4 sides + 4 corners = 6 ribs + 2
It therefore also applies to the body from which you started because of the previous operations, because all bodies are connected by actions that retain their validity.