Proof of the Series Representation of e...stuck Bottom Line: Prove that $e = 1+1+\frac{1}{2!}+\cdots$
Define $e = \lim\limits_{n \rightarrow \infty} \left(1+\frac{1}{n}\right)^n$
I would like to do it by expanding $\left(1+\frac{1}{n}\right)^n$ binomially as $\sum\limits_{j=0}^n {n \choose j}\left(\frac{1}{n}\right)^j$. Somehow I need to show that this approaches $\frac{1}{j!}$ as $n$ approaches $\infty$. That's basically it I think. I feel like I'm missing something easy, but I haven't been able to find this online except for a Dr. Math forum that I can't follow (here: Link). Please be thorough in your explanation, if possible.
Thanks for the help!
 A: If you want a maximum level of rigor, you can avoid the interchange of two limiting operator problem as follows:
As we observed,
$$ \left( 1 + \frac{1}{n} \right)^n = \sum_{j=0}^{n} \binom{n}{j}\frac{1}{n^j} = \sum_{j=0}^{n} \frac{1}{j!}\prod_{k=1}^{j} \left( 1 - \frac{k-1}{n} \right) $$
where the product is considered to yield $1$ if $j = 0$ by convention. Now, fix a positive integer $m$. Then by noting that each summand is non-negative, for any $n \geq m$ we have
\begin{align*}
\sum_{j=0}^{m} \frac{1}{j!}\prod_{k=1}^{j} \left( 1 - \frac{k-1}{n} \right)
&\leq \sum_{j=0}^{n} \frac{1}{j!}\prod_{k=1}^{j} \left( 1 - \frac{k-1}{n} \right)
 \leq \sum_{j=0}^{n} \frac{1}{j!}.
\end{align*}
This inequality shows that
\begin{align*}
\liminf_{n\to\infty} \sum_{j=0}^{m} \frac{1}{j!}\prod_{k=1}^{j} \left( 1 - \frac{k-1}{n} \right)
&\leq \liminf_{n\to\infty} \left(1 + \frac{1}{n}\right)^{n} \\
&\leq \limsup_{n\to\infty} \left(1 + \frac{1}{n}\right)^{n}
 \leq \limsup_{n\to\infty} \sum_{j=0}^{n} \frac{1}{j!}.
\end{align*}
But since
$$ \liminf_{n\to\infty} \sum_{j=0}^{m} \frac{1}{j!}\prod_{k=1}^{j} \left( 1 - \frac{k-1}{n} \right) = \sum_{j=0}^{m} \frac{1}{j!} $$
and
$$ \limsup_{n\to\infty} \sum_{j=0}^{n} \frac{1}{j!} = \sum_{j=0}^{\infty} \frac{1}{j!}, $$
it follows that
$$ \sum_{j=0}^{m} \frac{1}{j!}
\leq \liminf_{n\to\infty} \left(1 + \frac{1}{n}\right)^{n}
\leq \limsup_{n\to\infty} \left(1 + \frac{1}{n}\right)^{n}
\leq \sum_{j=0}^{\infty} \frac{1}{j!}. $$
Notice that this inequality holds for any $m$. Thus taking $m\to\infty$, we obtain the desired result.
A: As you point out, the desired formula follows without too much trouble if you can show that
$$
\lim_{n\to\infty} \binom nj \bigg(\frac{1}{n}\bigg)^j = \frac{1}{j!}.
$$
To see this, note that
\begin{align*}
\binom nj \bigg(\frac{1}{n}\bigg)^j &= \frac{n(n-1)(n-2)\cdots(n-(j-1))}{j!} \bigg(\frac{1}{n}\bigg)^j \\\
&= \frac{n(n-1)(n-2)\cdots(n-(j-1))}{n^j} \frac{1}{j!} \\\
&= 1\bigg(1-\frac1n\bigg)\bigg(1-\frac2n\bigg)\cdots\bigg(1-\frac{j-1}n\bigg) \frac{1}{j!}.
\end{align*}
Therefore
\begin{align*}
\lim_{n\to\infty} \binom nj \bigg(\frac{1}{n}\bigg)^j &= \lim_{n\to\infty} 1\bigg(1-\frac1n\bigg)\bigg(1-\frac2n\bigg)\cdots\bigg(1-\frac{j-1}n\bigg) \frac{1}{j!} \\\
&= \lim_{n\to\infty} \bigg(1-\frac1n\bigg) \lim_{n\to\infty} \bigg(1-\frac2n\bigg) \cdots \lim_{n\to\infty} \bigg(1-\frac{j-1}n\bigg) \cdot \frac1{j!} \\\
&= 1\cdot 1\cdots 1\cdot \frac1{j!} = \frac1{j!}.
\end{align*}
A: This might be cheating, but use:
$$f(x)=e^x = \lim\limits_{n \rightarrow \infty} \left(1+\frac{1}{n}\right)^{nx}$$
So that:
$$f^{(k)}(x) =  \lim\limits_{n \rightarrow \infty}  \left(1+\frac{1}{n}\right)^{nx} \left(\log\left(1+\frac{1}{n}\right)^n\right)^k = e^x$$
Now write the taylor series for $e^x$ using the above and plug in $x=1$.
