number $e$, why $\sum_{k=0}^\infty \frac{1}{k!}=e$? The definition that I have for $e$ is
$$e:=\lim_{n\to\infty }\left(1+\frac{1}{n}\right)^n.$$
How can we get that 
$$e=\sum_{k=0}^\infty \frac{1}{k!}\ \ ?$$
My try
By Newton formula
$$\left(1+\frac{1}{n}\right)^n=\sum_{k=0}^n\binom{n}{k}\frac{1}{n^k}=\sum_{k=0}^n\frac{n!}{k!(n-k)!}\frac{1}{n^k},$$
but I don't know how to prove that
$$\lim_{n\to\infty }\sum_{k=0}^n\frac{n!}{k!(n-k)!}\frac{1}{n^k}=\sum_{k=0}^\infty \frac{1}{k!}.$$
 A: As you say, by the Binomial Theorem,
$$
\left(1+\frac1n\right)^n
=\sum_{k=0}^n\binom{n}{k}\frac1{n^k}\tag{1}
$$
For each $k\le n$, by Bernoulli's Inequality,
$$
\begin{align}
\frac{\binom{n+1}{k}\frac1{(n+1)^k}}{\binom{n}{k}\frac1{n^k}}
&=\frac{n+1}{n-k+1}\left(1-\frac1{n+1}\right)^k\\
&\ge\frac{n+1}{n-k+1}\left(1-\frac{k}{n+1}\right)\\
&=1\tag{2}
\end{align}
$$
and for each $k\gt n$, $\binom{n}{k}\frac1{n^k}=0$. Thus, for each $k$,
$$
\binom{n+1}{k}\frac1{(n+1)^k}\ge\binom{n}{k}\frac1{n^k}\tag{3}
$$
Furthermore, since
$$
\binom{n}{k}\frac1{n^k}=\frac1{k!}\cdot\overbrace{\frac{n}{n}\cdot\frac{n-1}{n}\cdot\frac{n-2}{n}\cdots\frac{n-k+1}{n}}^{\text{$k$ terms}}\tag{4}
$$
we have, for $n\ge k$,
$$
\frac1{k!}\left(\frac{n-k+1}{n}\right)^k\le\binom{n}{k}\frac1{n^k}\le\frac1{k!}\tag{5}
$$
Thus, by the Squeeze Theorem,
$$
\lim_{n\to\infty}\binom{n}{k}\frac1{n^k}=\frac1{k!}\tag{6}
$$
So, by $(3)$ and $(6)$, each term of the sum in $(1)$ increases to $\frac1{k!}$. Therefore, by the Monotone Convergence Theorem,
$$
\begin{align}
\lim_{n\to\infty}\left(1+\frac1n\right)^n
&=\lim_{n\to\infty}\sum_{k=0}^n\binom{n}{k}\frac1{n^k}\\
&=\sum_{k=0}^\infty\frac1{k!}\tag{7}
\end{align}
$$
A: It's easier to use Taylor series at a=0 to get the function for e:
Taylor series is a series which is used to plot graphs,etc. and can be found using:
((x^n)f(a)[nth derivative of f(x) at x=a])/n!
When this term is added for n going from 0 to infinity you get the Taylor series and here since a=0(special form of Taylor Series called Maclaurin Series) all the derivatives and f(a) are 1, thus each term in the Taylor series reduces to (x^n)/n! . When this term is added from n=0 to n=infinity, you get the familiar result for e
