Euler's Number Identity In pre-calc, we learned that 
$$\sum_{k=0}^{\infty} \frac{1}{k!} = \lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{n}$$
I expanded the right hand side using Binomial Theorem. Here are my steps:
$$(x+y)^{n} = \sum_{k=0}^{n} {n \choose k} \cdot x^{n-k}\cdot y^{k}$$
$$x = 1, \space\space\space\space\space\space\space y = \frac{1}{n}$$
Plugging that in gives:
$$\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{n} = \lim_{n\to\infty}\left(\sum_{k=0}^{n} {n \choose k} \cdot \frac{1}{n^k}\right)$$
That means that
$$\sum_{k=0}^{\infty} \frac{1}{k!} = \lim_{n\to\infty}\left(\sum_{k=0}^{n} {n \choose k} \cdot y^{k}\right)$$
Could anyone help me finish this? Ideally, an explanation accessible to a middle school pre-calc student. Thanks. 
 A: $$\begin{pmatrix}n\\k\end{pmatrix}.\frac{1}{n^k}=\frac{n!}{n^kk!(n-k)!}$$
but I believe that you have got confused here. The easiest way to do it is:
$$\left(1+\frac1n\right)^n=1+n.\frac1n+\frac{n(n-1)}{2!\times n^2}+\frac{n(n-1)(n-2)}{3!\times n^3}...$$
and so we get that:
$$\lim_{n\to\infty}\left(1+\frac1n\right)^n=\lim_{n\to\infty}\left(1+n.\frac1n+\frac{n(n-1)}{2!\times n^2}+\frac{n(n-1)(n-2)}{3!\times n^3}...\right)=1+1+\frac{1}{2!}+\frac{1}{3!}...$$
$$=\sum_{n=0}^\infty\frac{1}{n!}$$
A: Hint: You can easily find the value of the limit using $\log$ technique. Also, you can find a solution using Taylor's series of $e^x$! To know more see here.
A: First note
$$\left(1+\frac1n\right)^n=\sum_{k=0}^n\frac{n!}{k!(n-k)!}\frac{1}{n^{n-k}}=\sum_{k=0}^n\frac{n(n-1)\cdots(n-k+1)}{n^{n-k}}\frac{1}{k!}\le\sum_{k=0}^n\frac{1}{k!}. $$
For the fixed $m<n$,
$$\left(1+\frac1n\right)^n=\sum_{k=0}^n\frac{n!}{k!(n-k)!}\frac{1}{n^{n-k}}>\sum_{k=0}^m\frac{n(n-1)\cdots(n-k+1)}{n^{n-k}}\frac{1}{k!}\ge\sum_{k=0}^m\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)\cdots\left(1-\frac{k-1}{n}\right)\frac{1}{k!}. $$
Let $n\to\infty$, one has
$$ \lim_{n\to\infty}\left(1+\frac1n\right)^n\ge\sum_{k=0}^m\frac{1}{k!}. $$
So for any $m$,
$$ \sum_{k=0}^m\frac{1}{k!}\le\lim_{n\to\infty}\left(1+\frac1n\right)^n\le e. $$
Let $m\to\infty$, one has
$$ \sum_{k=0}^\infty\frac{1}{k!}=\lim_{n\to\infty}\left(1+\frac1n\right)^n= e. $$
A: Let’s extend this to a more general form$$e^x=\lim\limits_{n\to\infty}\left(1+\frac xn\right)^n$$Expand the right - hand side with the binomial theorem to get that$$\begin{align*}e^x & =\lim\limits_{n\to\infty}\left[1+n\cdot\frac {x}{n}+\frac {n(n-1)}{2!}\left(\frac xn\right)^2+\frac {n(n-1)(n-2)}{3!}\left(\frac xn\right)^3+\cdots\right]\\ & =\lim\limits_{n\to\infty}\left[1+x+\frac {(n^2-n)x^2}{2!n^2}+\frac {(n^3-3n^2+2n)x^3}{3!n^3}+\cdots\right]\\ & =1+x+\frac {x^2}{2!}+\frac {x^3}{3!}+\cdots\end{align*}$$Therefore$$e^x\color{blue}{=\sum\limits_{n\geq0}\frac {x^n}{n!}}$$
