Arguing inductively, it suffices to prove that if with separable elements over , then is simple. We also assume that is infinite, since the finite case is covered by Finite extension of a Galois field.

Suppose we have distinct morphisms of field extensions . Then the polynomials

are distinct, for otherwise and implying . Therefore the polynomial

is not identically zero. Since is infinite, it follows that there exists a so that the evaluation . This means distinct map to distinct elements . Since the cardinality of is the separable degree , and each is a root of the minimal polynomial of over , we have

But by hypothesis the extension is separable, so the upper and lower bounds are equal, squeezing , whence .