Deriving the power series for $e$ simply? The definition of $e$ is:
$$e = \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n$$
If we use the Binomial Theorem on the function itself:
$$\left(1 + \frac{1}{n}\right)^n = \sum_{k=0}^{n} \binom{n}{k}\frac{1}{n^k} = \sum_{k=0}^{n} \frac{n!}{k!(n-k)! \cdot n^k}$$
At this point I am not sure where to go, how to both manipulate this expression and also have it be a valid input to the limit definition. Is there a simple approach that does not require Calc II knowledge, or is this one of those problems where there's really just one way to do it that everyone uses?
 A: Rudin has a nice proof of this fact (Principles of mathematical analysis, p65)

A: I think you are trying to show
$$\lim_{n \to \infty} (1+1/n)^n = \sum_{k=0}^\infty \frac{1}{k!}.$$
That's the same as showing that
$$\lim_{n \to \infty} \sum_{k=0}^n \frac{n!}{k!(n-k)! n^k}=\sum_{k=0}^\infty \frac{1}{k!}$$
by the binomial theorem as you have shown. You can also notice that
$$\lim_{n \to \infty} \frac{n!}{(n-k)! n^k}=1$$
for any fixed $k$. This suggests but does not prove the result. Generally there are theorems giving sufficient conditions for interchange of limit and summation, but you are not likely to know any of them at your stage. 
One way to prove this result directly is to bound the left side from above and below by the same number. The "above" part is easy:
$$(1+1/n)^n \leq \sum_{k=0}^n \frac{1}{k!}$$
which is easy to see from the binomial theorem. Taking limits on both sides gives the bound we want. On the other hand
$$(1+1/n)^n \geq C_{m,n} \sum_{k=0}^m \frac{1}{k!}$$
if $n \geq m$, where 
$$C_{m,n}=\prod_{i=1}^{m-1} \left ( 1-\frac{i}{n} \right ).$$
(This step requires some fiddling with factorials that I don't want to write out.)
Therefore, for all $m$
$$\lim_{n \to \infty} (1+1/n)^n \geq \sum_{k=0}^m \frac{1}{k!}$$
because $\lim_{n \to \infty} C_{m,n}=1$. But now the left side doesn't depend on $m$, so we can send $m \to \infty$ to get
$$\lim_{n \to \infty} (1+1/n)^n \geq \lim_{m \to \infty} \sum_{k=0}^m \frac{1}{k!}.$$
