Let f be analytic on a region. Remember that the nth derivative can be expressed as a contour integral. specifically, the nth derivative at p is n!/2πi times the integral of f(z)/(z-p)n+1, using any contour that surrounds p.
Draw a circle of radius r around p. Let m be the maximum of |f| on the disk of radius r. The integrand is bounded by m/rn+1. The directional derivative around the circle has norm r, and θ runs from 0 to 2π. Put it all together and the nth derivative is bounded by mn!/rn.
Let f be entire, i.e. everywhere analytic, and suppose f is bounded by m. Let p be any point in the complex plane. What is f′(p)? Its norm is bounded by something over r, and r can be arbitrarily large, hence the first derivative is 0. All derivatives are 0 everywhere, and f is constant.
If the real part of an analytic function f is bounded, Ef is analytic and bounded. Thus Ef is constant, and f is constant. If the imaginary part is bounded, use Eif.