Here's a nice theorem by Gauss
The (2D) shoelace theorem
Suppose $\mathbf{x_1},\mathbf{x_2},...,\mathbf{x_n}$ are ordered vertices of a polygon. Let $x_i, y_i$ be the coordinates of $\mathbf{x_i}$ and define $\mathbf{x_{n+1}}$ to be $\mathbf{x_1}$. Then the area of the polygon is plus or minus
$$
\frac{1}{2}\sum_i{x_iy_{i+1} - x_{i+1}y_i}
$$
This formula gives rise to the name of the theorem as shown by the illustration below
Proof
We're going start off by proving the simple case where the polygon is a triangle. The first thing to notice is that if you take any two adjacent vertices, .e.g. $\mathbf{x_1}$ and $\mathbf{x_3}$, they form a triangle with the origin. This has an area equal to half of the parallelpiped formed by the two vertices (treated as vectors from the origin). And the area of the parallelpiped is plus or minus $det(\mathbf{x_1}\vert \mathbf{x_3})$, with the sign being positive if the columns taken in order produce an anti-clockwise motion around the origin.
If you go anti-clockwise around the triangle and add together all the determinants (multiplied by a half) you get the area of the triangle, as shown in the diagram above. This produces the shoelace formula given by Gauss.
The same argument can be used for any polygon. However, if you don't find that intuitive you can imagine the polygon divided into triangles. If you sum the areas of the triangles the values for the internal edges cancel out. This is because the edge appears in two determinants, but with a different column ordering in each matrix.
There are other ways to prove the shoelace theorem, but the nice thing about this version of the proof is that it extends naturally to three or more dimensions....
The 3D shoelace theorem
If a polyhedra has faces $1,...n$ and face $i$ has vertices $\mathbf{x_{i1}},\mathbf{x_{i2}},\mathbf{x_{i3}}$ then the volume of the polyhedra is plus or minus
$$
\frac{1}{6}\sum{det(\mathbf{x_{i1}}\vert \mathbf{x_{i2}}\vert \mathbf{x_{i3}})}
$$
However, this only works if edge vertices appear in a different order, modulo rotation, in each of the two faces in which they appear. For example if one face is $\mathbf{a},\mathbf{b},\mathbf{c}$, then an adjacent face may be $\mathbf{b},\mathbf{a},\mathbf{d}$, or even $\mathbf{a},\mathbf{d},\mathbf{b}$, but not $\mathbf{a},\mathbf{b},\mathbf{d}$.
Proof
The proof is identical to the 2D case. The only part that differs is that the constant at the front is $\frac{1}{6}$ instead of $\frac{1}{2}$. What we need to show, therefore, is that the irregular tetrahedron, formed by $\mathbf{x_{i1}},\mathbf{x_{i2}},\mathbf{x_{i3}}$ and the origin, has an area equal to $\frac{1}{6}$ of $det(\mathbf{x_{i1}}\vert \mathbf{x_{i2}}\vert \mathbf{x_{i3}})$.
The following diagram shows this is true for special case $\mathbf{x_{i1}}=\mathbf{\hat{i}},\mathbf{x_{i2}}=\mathbf{\hat{j}},\mathbf{x_{i3}}=\mathbf{\hat{k}}$. It uses the formula for the volume of a pyramid $V = \frac{1}{3} \times \text{height} \times \text{base_area}$.
The higher dimensional shoelace theorem
I haven't proven this, but the result seems intuitive and the 3D proof gives a sketch of how to proceed
If an N dimensional polytope has sides $1,...n$ and
face $i$ has vertices $\mathbf{x_{i1}},\mathbf{x_{i2}},...,\mathbf{x}_{\text{iN}}$ then the volume of the polytope is plus or minus
$$
\frac{1}{N!}\sum{det(\mathbf{x_{i1}}\vert \mathbf{x_{i2}}\vert...\vert \mathbf{x_{in}})}
$$
However, this only works if for every pair of neighbouring N-1 dimensional faces, the vertex order given for one is an odd permutation of the vertex order given for the other. (Where the unshared vertex in the first face is treated as being identical to the unshared vertex in the second face for the purpose of identifying the permutation).
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