Given that $\lim\limits_{n \to \infty}\left(1+\frac{1}{n}\right)^n=e$, show that $1+\frac{1}{1!}+\frac{1}{2!}+\cdots=e$. Problem
Given that $\lim\limits_{n \to \infty}\left(1+\dfrac{1}{n} \right)^n = e$, show that $1+\dfrac{1}{1!}+\dfrac{1}{2!}+\cdots=e$.
Proof
By binomial theorem, we have 
\begin{align*}\left(1+\dfrac{1}{n}\right)^n&=\sum_{k=0}^{k=n}\binom{n}{k}1^k \left(\frac{1}{n}\right)^{n-k}\\
&=\sum_{k=0}^{k=n}\frac{1}{k!}\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)\cdots \left(1-\frac{k-1}{n}\right).
\end{align*}
Thus, on one hand, $$\left(1+\dfrac{1}{n}\right)^n \leq \sum_{k=0}^{k=n}\frac{1}{k!}.\tag1$$
Take the limits as $n \to \infty$ on the both sides of $(1)$. We have $$e=\lim_{n \to \infty}\left(1+\dfrac{1}{n}\right)^n\leq \varliminf_{n \to \infty} \sum_{k=0}^{k=n}\frac{1}{k!}.\tag2$$
On the other hand, take a positive integer $m$ such that $m<n$ and fix it. We have 
$$\left(1+\dfrac{1}{n}\right)^n \geq \sum_{k=0}^{k=m}\frac{1}{k!}\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)\cdots \left(1-\frac{k-1}{n}\right).\tag3$$
Likewise,take the limits as $n \to \infty$ on the both sides of $(3)$. We have 
$$e=\lim_{n \to \infty}\left(1+\dfrac{1}{n}\right)^n\geq \sum_{k=0}^{k=m}\frac{1}{k!}.\tag4$$
Take the limits as $m \to \infty$ on the both sides of $(4)$. We have 
$$e \geq \varlimsup_{m \to \infty}\sum_{k=0}^{k=m}\frac{1}{k!}.\tag 5$$
Combine $(2)$ and $(5)$. It follows that $$\lim_{n \to \infty}\sum_{k=0}^{k=n}\frac{1}{k!}=e.$$
Please correct me If I'm faulty. Hope to see other solutions.
 A: Another more elegant proof
At the beginning, let's recall an unusual fact, named Tannery's limit theorem, which states that

Let $S(n)=\sum\limits_{k=0}^{n}a_k(n)$.
If the following conditions are all satisfied:

*

*For any $k$, $\lim\limits_{n \to \infty}a_k(n)=a_k$;

*for any $k$，there exists a $M_k>0$ independent of $n$ such that     $|a_k(n)|\leq M_k$;

*$\sum\limits_{k=0}^{\infty}M_k<\infty$,

then $$\lim\limits_{n \to\infty}S(n)=\sum\limits_{k=1}^{\infty}a_k.$$

As for the present problem, we may denote $$S(n)=\left(1+\frac{1}{n}\right)^n,$$ and$$a_k(n)=\frac{1}{k!}\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)\cdots \left(1-\frac{k-1}{n}\right).$$ Thus, we may verify that:

*

*For any $k$, $\lim\limits_{n \to \infty}a_k(n)=\dfrac{1}{k!}=a_k$;

*for any $k$，$|a_k(n)|\leq \dfrac{1}{k!}=M_k$;

*$\sum\limits_{k=0}^{\infty}\dfrac{1}{k!}=\sum\limits_{k=0}^{\infty}M_k<2+\sum\limits_{k=2}^{\infty}\dfrac{1}{2^{k-1}}=3<\infty$.

It follows that $$e=\lim\limits_{n \to \infty}\left(1+\frac{1}{n}\right)^n=\lim\limits_{n \to \infty}S(n)=\sum\limits_{k=0}^{\infty}\dfrac{1}{k!}=\sum\limits_{k=0}^{\infty}a_k.$$
A: \begin{eqnarray*}
 \lim_{n\to \infty} \left(1+\frac{1}{n}\right)^{n} &=&  \lim_{n\to \infty}  \sum_{k=0}^{n}{\binom{n}{k} \left(\frac{1}{n}\right)^{k}} \\
&=&  \lim_{n\to \infty}  \sum_{k=0}^{n}\frac{1}{k!}{\prod_{m=0}^{k-1}{\left(1-\frac{m}{n}\right)} } \\
&=& \sum_{k=0}^{\infty}{\frac{1}{k!}\prod_{m=0}^{k-1}{ \lim_{n\to \infty} \left(1-\frac{m}{n}\right)} }
\end{eqnarray*}
The key thing is that the  limit of $\prod_{m=0}^{k-1}{\left(1-\frac{m}{n}\right)}$ as $n$ grows without bound becomes $1$.
A: The idea of this approach is to define exponential function $\exp x$ by series, and then prove that $\exp 1 = e$. However, we use quite a few of standard properties of $\exp$ and $\ln$, so potential caveat is that some of these properties might not be easily proven by using such definition of $\exp x$, but I was not able identify any such weak part (required properties are listed below, scrutiny is welcome).

Definition 1. $e:=\lim\limits_{n \to \infty}\left(1+\frac{1}{n}\right)^n$.
Definition 2. $\exp(x):=\sum_{n\geq0}\frac{x^n}{n!}$.

Some useful standard results:


*

*$\exp x$ is strictly monotone (increasing) on $\mathbb{R}$,

*$\exp 0 = 1$ (follows from definition)


Then we are justified in following definition:

Definition 3. Let $\ln x$ be an inverse of $\exp(x)$, i.e. $\ln(\exp(x))=\exp(\ln(x))=x$.

For the proof we will require also these properties: 


*

*$\ln x$ is continuous,

*$\ln x$ is strictly increasing,

*$\ln x^n=n \ln x$.

*$\frac{d}{dx} \ln x=\frac{1}{x}$. (follows from derivative of inverse function and from $(\exp x)'=\exp x$, which can be shown from definition)

*$\ln 1 = 0$ (follows from $\exp 0 = 1$ and inverse)


Let's first show that $\ln x=1$ has unique solution and this solution is $e$ (proof inspired by  https://proofwiki.org/wiki/Euler%27s_Number:_Limit_of_Sequence_implies_Base_of_Logarithm). Indeed we  have
\begin{align}
\ln e &= \ln\left(\lim_{n \to \infty} \left(1+\frac{1}{n}\right)^n\right)\\
&= \lim_{n \to \infty}\left(\ln \left(1+\frac{1}{n}\right)^n\right)\tag{1}\\
&= \lim_{n \to \infty}\frac{\ln \left(1+\frac{1}{n}\right)}{1/n}\\
&= \lim_{h \to 0}\frac{\ln \left(1+h\right)-\ln 1}{h}\\
&= \frac{d}{dx}\ln x|_{x=1}\tag{2}\\
&= \frac{1}{x}|_{x=1}\\
&= 1\\
\end{align}
where in $(1)$ we used that $\ln x$ is continuous, and in $(2)$ we have used the definition of derivative. Since the $\ln x$ is injective, $e$ is the unique solution. 
Now it follows
$$e = \exp (\ln e) = \exp(1) =\sum_{n\geq0}\frac{1}{n!},$$
which is what we wanted to prove.
