But it isn't that hard to prove for yourself for the simple case of the binomial curve if you're willing to supply some elbow grease. Here is an outline of how I did it when I was an undergrad.
1. Prove that log(n!) = n log(n) - log(n)/2 + k + o(1) where k is some constant. (Trick: approximate the integral of log(x) with the trapezoidal rule, that gives you n log(n). You wind up with log(n)/2 + log(1)/2 left over. Prove that the sum of the errors to infinity converges to a constant, which gives you, to finite n, k + o(1) where the o(1) term is the sum of the rest of the errors that you haven't reached yet.)
2. Suppose that X_1, X_2, etc are an identically distributed independent set of random variables that have probability 0.5 of being +1 or -1. Prove that each X_i has mean 0 and variance 1.
3. Consider the random variable Y_n = (X_1 + X_2 + ... + X_n)/sqrt(n). Prove that it has mean 0 and variance 1.
4. What is the "density" (in quotes because it is discrete) of Y_n around x? Well you can calculate an exact formula in terms of the binomial formula, and stick in your approximation from step 1, then show that the point weight of the nearest point to x is going to be e^(-x^2)(c + o(1))/sqrt(n) where c is a constant that can be derived from k in step 1. Remember that that's the only point for a distance of sqrt(n), so the density is proportional to e^(-x^2).
5. Look up the standard argument that the integral from -infinity to infinity of e^(-x^2) is sqrt(pi). Note that Y has to sum up to a total weight of 1 (because it is a probability density), substitute that fact in then you get both the standard normal distribution out AND as an unexpected bonus, Stirling's formula!
(If you wish you can do the same for a biased coin, the derivation is slightly more complex but follows the same lines. However to get arbitrary probability distributions you need to use different approaches. People usually don't encounter that proof until graduate school.)