Where does the guassian function/normal or bell curve come from? I am confused as to where the function for the normal distribtuion comes from.  Where does the e and pi come from?  In my textbook I am presented with the function,but I am unsure about where it came from.  I have spent hours on google and have yet to find a good proof that i can understand.  Come someone provide a source that shows an elementary proof of the normal curve function.
The Gaussian (normal or bell curve) arises, when you look at a huge collection of random variables and whose distribution you are not aware of.

You might want to look up Central Limit Theorem (CLT). The CLT explains why when you have a huge collections of random variables, they appear to follows the Gaussian.

A version of the CLT goes as follows.

Consider $n$ independently and identically distributed random variables, $\{X_1,X_2,\ldots,X_n\}$, with mean $\mathbb{E}(X_k) = \mu$ and variance $\mathbb{E}((X_k-\mu)^2) = \sigma^2 &lt; \infty$. Now look at the sample mean $$S_n = \dfrac{X_1 + X_2 + \cdots + X_n}{n}$$
The central limit theorem states that $$\sqrt{n}(S_n - \mu) \to_{d} \mathcal{N}(0,\sigma^2)$$ as $n \to \infty$.

A proof for a weak version goes like this. Consider the characteristic function of the random variable $X$, say $\phi_X(t)$. Expanding using Taylor series about $0$, we get that $$\phi_{X-\mu}(t) = \phi_{X-\mu}(0) + \phi_{X-\mu}'(0) t + \dfrac{\phi_{X-\mu}''(0)}2 t^2 + \mathcal{O}(t^4/n^4) = 1 - \dfrac{\sigma^2}2 t^2 + \mathcal{O}(t^4/n^4)$$
The characteristic function for $(S_n - \mu)$ is $$\phi_{S_n - \mu}(t) = \phi_{X - \mu}(t/n)^n = \left( 1 - \dfrac{\sigma^2}2 t^2/n^2 + \mathcal{O}(t^4/n^4)\right)^n$$
Hence, the characteristic function for $Z_n = \sqrt{n}(S_n - \mu)$ is $$\phi_{Z_n}(t) = \phi_{\sqrt{n}(S_n - \mu)}(t) = \phi_{S_n - \mu}(\sqrt{n}t) = \left( 1 - \dfrac{\sigma^2}2 nt^2/n^2 + \mathcal{O}(t^4/n^2)\right)^n = \left( 1 - \dfrac{\sigma^2 t^2/2}n + \mathcal{O}(t^4/n^2)\right)^n$$
Hence, $$\lim_{n \to \infty} \phi_{Z_n}(t) = \lim_{n \to \infty} \left( 1 - \dfrac{\sigma^2 t^2/2}n + \mathcal{O}(t^4/n^2)\right)^n = \exp(-\sigma^2 t^2/2)$$

Note that the proof above is true for independent, identically distributed random variables arising from any nice probability distribution. With a little more effort, you can show that the convergence holds for independent (but not necessarily identically distributed) random variables. This is why the normal distribution is fundamental and appears everywhere in probability and statistics. The number $e$ appears since $$e = \lim_{n \to \infty} \left(1+\dfrac1n \right)^n.$$ The $\pi$ you are reffering to appears as a normalizing constant since $$\int_{-\infty}^{\infty} \exp(-x^2) dx = \sqrt{\pi}$$
I'm not aware of an elementary derivation of the functional form itself, but I can give you my favorite bit of intuition on where it comes from.

Suppose I told you I had a function $f : \mathbb{R} \to \mathbb{R}$, and that $f(0) = 0$.  Based on this extremely limited information, I then ask you to find me $g(a \, | \, x)$, a probability density function for the value of $f(x)$.

There is one crucial observation you can make that allows you to find a partial solution.  That observation is: for any $a, \Delta \in \mathbb{R}$, intuitively speaking, it must be true that $P(f(x) = a \, | \, f(0) = 0) = P(f(x + \Delta) = b + a \, | \, f(\Delta) = b)$.  In other words, you have no reason to believe that the point $(0, 0)$ is special: you might as well rephrase the problem as what are the odds that $f$ increases by $a$ when its parameter is changed by $\Delta$? and assume that these odds are the same at any point of $f$.

We can use this to set up an integral equation.  When $f$ travels from $(0, 0)$ to $(x, ?)$, imagine it first travelling to some intermediate point $x'$ and then on to $x$ from there.  Then for any $x'$ between $0$ and $x$, it must be true that:
$$g(a \, | \, x) = \int_{-\infty}^{\infty} \, g(a' \, | \, x') g(a - a' | x - x') \, da'$$

The solution to this integral equation is:
$$\frac{1}{\sqrt{2\pi c_1x}} \, e^{-\frac{1}{2}\frac{(a - c_2 x)^2}{c_1 x}}$$

This equation is commonly referred to as the Gaussian with drift.  $c_1 x$ is the variance and $c_2 x$ is the mean.