Fix any nonzero a in I. Consider the map f: I --> I given by f(x) = xa. Note that f is a left R-module homomorphism and hence the image and kernel are left ideals of R. Since I is minimal, it follows that either f is zero or an isomorphism. If f is zero for every a in I, then I² = 0. Otherwise, there is a nonzero a in I such that f is an isomorphism. So there exists a nonzero e in I such that ea = a. Thus, e²a = ea and hence (e² - e)a = 0. This means that e² - e is in the kernel of f. But f is an isomorphism, so e² - e = 0. Whence e is idempotent. Moreover, since e is nonzero it generates I. (Note that I is a simple left R-module since it is a minimal left ideal of R. A simple module is generated by any one of its nonzero elements.)