Despite the funny name this is a relatively important finding from linear algebra*, and it is exceedingly simple to use and prove. It takes two square matrices A and B, which have n rows and n columns. If these two matrices are invertible, i.e. their rank = n, then we know this about the inverse of their product: (AB)-1 = B-1A-1.
This finding is referred to as the shoe-sock theorem because it resembles the steps needed to invert (that is, undo) the action of putting on socks and shoes. If A is for socks, B is for shoes, and -1 is the process of taking them off, to take them both off (AB)-1 one must first take off the shoes B-1, then the socks A-1. Hey, I said it was a funny name, not a particularly insightful one...
One proof is nice and easy, though I'm sure harder ones can be devised without much difficulty. This proof relies on the fact that we need to prove B-1A-1 to be the inverse of AB. We'll do this by showing
(B-1A-1)(AB) = In
(AB)(B-1A-1) = In.
In other words, we'll show that multiplying AB on either side will produce the identity matrix of size n (that is, In), which is necessary and sufficient proof that the inverse of AB is B-1A-1.
First: (B-1A-1)(AB) = B-1(A-1A)B = B-1InB = B-1B = In -- cool.
And also: (AB)(B-1A-1) = A(BB-1)A-1 = AInA-1 = AA-1 = In -- very cool, statement proven.
*User zfcorey reminded me that this rule is true of any abstract algebra with an identity.