How does $\sum_{n=0}^{\infty}\frac{p^n}{n!} = e^p$? How does $\sum_{n=0}^{\infty}\frac{p^n}{n!} = e^p$? I know that it is a geometric sum but I can't seem to work it out. 

 A: A very polished development of this series expression for the exponential function is given in the opening section of W. Rudin's Real and Complex Analysis.  Rudin's approach is to take the power series as defining a holomorphic function (since the radius of convergence is infinite) and briefly prove that the function equals its derivative.  Moreover if one specifies the value at the origin $f(0)=1$, then this characterizes the exponential function.  The usual properties of the exponential function can also be deduced from the series, e.g. $e^x e^{-x} = 1$, from which it follows that $e^x$ is never zero.
A: Richard, the sum is not a geometric series, since the ratio is not constant:
$$\frac{a_{n+1}}{a_n}=\frac p {n+1} $$ It is rather the series representation for the exponential function $y=e^x$.
We know from the theory of Taylor polynomials and series that if we have a function $f$ that is differentiable inifinitely many times, then under appropriate conditions we can define it's Taylor series around a point $x=a$ as
$$f(x)=\sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n$$
The appropriate condition is that the error produced by the $n$th approximation, which assuming the integrability of $f^{(n)}$, can be proven to be
$$E_n(a,x) = \int_a^x\frac{(x-t)^n}{n!}f^{(n+1)}(t)dt$$
will tend to zero for $n\to \infty$. The error expression which looks quite strange is not too hard to derive. I transcribe from this question of mine:

$$f(x)=f(a)+f'(a)(x-a)+R_1(x)$$ so that
$$R_1(x) = f(x)-f(a) - f'(a) (x-a)$$
$${R_1}(x) = \int\limits_a^x {f'\left( t \right)dt}  - \int\limits_a^x {f'(a)dt} $$
$${R_1}(x) = \int\limits_a^x {f'\left( t \right) - f'\left( a \right)dt} $$
So now we integrate by parts with
$$f'\left( t \right) - f'\left( a \right) = u$$
$$t - x = v$$
to get
$${R_1}(x) = \int\limits_a^x {\left( {x - t} \right)f''\left( t \right)dt} $$
We can similarily do this with $R_2(x)$, since
$${R_2}(x) = {R_1}(x) - f''\left( a \right)\frac{{{{\left( {x - a} \right)}^2}}}{{2!}}$$
$${R_2}(x) = \int\limits_a^x {\left( {x - t} \right)f''\left( t \right)dt}  - \int\limits_a^x {\left( {x - t} \right)f''\left( a \right)dt} $$
$${R_2}(x) = \int\limits_a^x {\left( {x - t} \right)\left( {f''\left( t \right) - f''\left( a \right)} \right)dt} $$
So again integrating by parts gives
$${R_2}(x) = \int\limits_a^x {\frac{{{{\left( {x - t} \right)}^2}}}{{2!}}f'''\left( t \right)dt} $$

In the case of the exponential function, we can prove that the error indeed goes to zero, from where we can represent it around $x=a$ as
$$\exp(x)=\sum_{n=0}^{\infty} \frac{e^a }{n!}(x-a)^n$$
since it is the case $f^{(n)}(a) = e^a$ for the exponential function.
It is important to note some authors define $$\exp(x):=\sum_{n=0}^{\infty} \frac{x^n }{n!}$$
and derive then the properties of the function (note this is the series around $x=0$).
NOTE The error term in $(1)$ is used whenever $f^{(n)}$ is integrable. Else, we need other formulas, such as Cauchy's or Lagrange's.
