Why does the sum of the reciprocals of factorials converge to $e$? I've been asked by some schoolmates why we have $$ \sum_{n=0}^\infty \frac{1}{n!}=e.$$ I couldn't say much besides that the $\Gamma$ function, analytic continuation of the factorial, is defined with an integral involving $e$. Then I also know that actually $$ \sum_{n=0}^\infty \frac{x^n}{n!}=e^x.$$ Is there a reason for these facts?
P.S. I added the tag "intuition", please remove it if you think it is not pertinent.
 A: Since you are calling for intuition, a term by term version. Intuitively, the inverse factorial coefficient $\frac{1}{n!}$ is the most natural, as it yields "some kind of invariance to differentiation", since:
$$\left(\frac{x^n}{n!}\right)' = \left(n\frac{x^{n-1}}{n!}\right) =  \frac{x^{n-1}}{(n-1)!} = \frac{x^{m}}{m!}$$
with $m= n-1$.
You may know that, and start with, a fundamental property of the natural exponential:  it is equal to its derivative. 
So, suppose that
$$ f(x) =  \sum_{n=0}^\infty a_n{x^n}\,,$$
is equal to its derivative. Then formally (I am skipping issues on convergence), 
$$ f'(x) =  \sum_{n=1}^\infty n a_n{x^{n-1}}\,,$$
thus by reindexing:
$$ f'(x) =  \sum_{n=0}^\infty (n+1) a_{n+1}{x^{n}}\,,$$
then, one should have, term by term:
$$ a_n= (n+1) a_{n+1}\,,$$
hence
$$  a_{n+1} = \frac{a_n}{n+1} =\frac{a_0}{(n+1)!}\,.$$
Since the exponential is the reciprocal to the $\log$, you require that $f(0)=1$, hence $a_0=1$. So naturally, 
$$ f(x) =  \sum_{n=0}^\infty \frac{x^n}{n!}=e^x\,,$$
and 
$$ f(1) =  \sum_{n=0}^\infty \frac{1}{n!}=e\,.$$
A: Here is a complete rigorous proof, not using power series/exponential function. Suppose we are just learning about limits and series - we didn't learn about uniform convergence, etc.
Take this as the definition of $e$:  $e = \lim_{n \to \infty} (1 + \frac 1 n)^n$. Then it is trivial to show that for any FIXED $N$, no matter how large (but FIXED), we also have $\lim_{n\to\infty}(1 + \frac 1 n)^{n+N} = e$ as well.
Let $S_n = \sum_{k=0}^n\frac 1 {k!}$. By the binomial theorem, $(1+\frac 1 n)^n \le S_n$ for all $n$, so we get that the sum of the infinite series is at least $e$ (comparison term by term: ${n \choose k} \cdot \frac 1 {n^k} \le \frac 1 {k!}$ for all $k$ from $0$ to $n$). This is the easy part.
For the other part, FIX a natural number N (however large, but FIXED). Again by the binomial theorem, and through the same kind of computation, we have $S_N \le (1+\frac 1 n)^{n+N}$. We only need the first $N+1$ terms of the expanded sum from the binomial theorem; $\frac 1 {k!} \le { n+N \choose k} \cdot \frac 1 {n^k}$. Now keep $N$ fixed and let $n \to \infty$; this shows that $S_N \le e$. Finally, since this is true for EVERY $N$, we get the opposite inequality: the sum of the series of inverses of factorials is $\le e$. By the way, this also shows the series converges (which we could see in other ways, but this by itself is a complete proof - we showed the partial sums are bounded by $e$).
A: Here is one approach.
Let $b > 1$.  If you compute the derivative of the function $b^x$, you find that the answer is just $b^x$ multiplied by an (annoying) constant.
There is a value of $b$ for which this constant is equal to $1$.  That's nice!  With this special value of $b$, the derivative of $b^x$ is just $b^x$, the same thing we started with.  That's a very neat property for a function to have.
This special value of $b$ is $e = 2.718 \ldots$.
It is now easy to compute the Taylor series of the function $e^x$ (centered at $0$).  We find that
\begin{equation}
\tag{$\spadesuit$} e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots.
\end{equation}
This comes directly from the Taylor series formula
\begin{equation}
f(x) = f(x_0) + f'(x_0)(x - x_0) + \frac{f''(x_0)}{2!}(x - x_0)^2 + \cdots.
\end{equation}
Plugging $x = 1$ into ($\spadesuit$) yields
\begin{equation}
e = \sum_{n=0}^{\infty} \frac{1}{n!}.
\end{equation}
A: By definition,
$$e=\lim_{n\to\infty}\left(1+\frac1n\right)^n.$$
Using the binomial theorem, the $k^{th}$ term of the development  is
$${\binom nk}\frac1{n^k}=\frac{n(n-1)(n-2)\dots(n-k+1)}{k!.n.n.n\dots n},$$
and 
$$\lim_{n\to\infty}{\binom nk}\frac1{n^k}=\frac1{k!}.$$
For example, 
$$\left(1+\frac1{1000}\right)^{1000}=\frac1{0!}+\frac1{1!}+\frac{0.999}{2!}+\frac{0.997002}{3!}+\frac{0.994010994}{4!}\dots$$
A: When you multiply $\exp(x)$ by $\exp(y)$ by that definition, you get $\exp(x+y)$.  That is one of the exponent laws, and is why $\exp(x)=e^x$ for some number $e$.  Then $e^1=\exp(1)$ which is your sum.
$$\exp(x)\exp(y)=\sum_{n=0}^{\infty}\frac{x^n}{n!}\sum_{m=0}^{\infty}\frac{y^m}{m!}\\
=\sum_{n=0}^{\infty}\sum_{m=0}^{\infty}\frac{x^ny^m}{n!m!}\\
=\sum_{n=0}^{\infty}\sum_{m=0}^{\infty}\frac{(m+n)!}{m!n!}\frac{x^ny^m}{(m+n)!}\\
\text{Let $k=n+m$.  Then sum along diagonals of constant $k$.}\\
=\sum_{k=0}^{\infty}\frac1{k!}\sum_{n=0}^{k}{k\choose n}x^ny^{k-n}\\
=\sum_{k=0}^{\infty}\frac1{k!}(x+y)^k\\=\exp(x+y)
$$
A: Let $$I_n=\int_1^e\frac{(\ln x)^n}{x^2}dx.$$ Using integration by parts we have $$I_{n+1}=-\frac{1}{e}+(n+1)I_n.$$
Using induction we can prove that
$$\frac{I_n}{n!}=1-\frac{1}{e}\sum_{k=0}^n \frac{1}{k!}$$
since $(I_n)_n$ is bounded, we let $n\to \infty$ we get
$$\sum_{k=0}^\infty \frac{1}{k!}=e.$$
A: These two familiar sums are the Taylor series for $e^x$ about $0$.
To get $e$ itself, you evaluate this series at $x=1.$
Derivation: The $n$th term of the Taylor series of a function $f$ about $a$ is
$$ \frac{f^{(n)}(a)}{n!} (x-a)^n.$$
But if $f(x) \triangleq e^x$, then $f'(x) = e^x$, and by an inductive argument,
$f^{(n)}(x) = e^x$ for every positive integer $n.$
Taking the series about $a = 0,$ the $n$th term is
$$ \frac{f^{(n)}(a)}{n!} (x-a)^n = \frac{e^a}{n!} (x-a)^n
 = \frac{e^0}{n!} (x-0)^n = \frac{x^n}{n!}.$$
That is, the Taylor series of $e^x$ as a function of $x$ about $0$ is
$$ e^x = \sum_{n=0}^\infty \frac{x^n}{n!}, $$
and by setting $x=1$ we get
$$ e = \sum_{n=0}^\infty \frac{1}{n!}, $$
A: Let $x \in \mathbb{R}^+$.
$f:x\longrightarrow \exp(x) \in C^\infty(\mathbb{R}, \mathbb{R})$, hence we can write $\forall n \in \mathbb{N}$ :
$$\left|f(x)-\sum_{k=0}^n f^{(k)}(0)\frac{x^k}{k!}\right| \leq \frac{|x|^{n+1}}{(n+1)!}I_{n+1}  \text{ where } I_{n+1}=sup_{[0,x]}|f^{n+1}|$$
And $\forall t \in [O,x] f^{(n)}(t)=\exp(t)$ which means $I_{n+1}=\exp(x)$
But $\displaystyle\lim _{n\to \infty }\left(\frac{|x|^{n+1}}{(n+1)!}\exp(x)\right)=\:0$
So $\displaystyle\lim _{n\to \infty }\left(\sum_{k=0}^n\frac{x^{k}}{k!}\right)=\exp(x)$
Just take $x = 1$ for $\mathbb{e}$
A: By definition 
$$\frac{d}{dx}e^x = e^x$$
therefore
$$\int_0^xe^t*1dt=e^x-1$$
we can use integration by parts which state that:
$$\int{f(x)g(x)dx}=f(x)\int g(x) - f^`(x)\int\int g(x)+...$$
let f(x)= e^x, g(x) = 1
$$e^x\int1-e^x\int\int1+...=e^x-1$$
we simplify to get:
$$\frac{1}{e^x}=1-\int1+\int\int1-\int\int\int1-...$$
we know that $\int\int..n..\int1dx=\frac{x^n}{n!}$ (basic integration rule) 
therefore
$$\frac{1}{e^x}=1-\frac{x^1}{1!}+\frac{x^2}{2!}-...$$

let substitute x = -n, we know that (-x)^n = -x^n when n is odd
$$e^n = 1+\frac{x^1}{1!}+\frac{x^2}{2!}+..$$
in fact, if we are to substitute x = ln(f(n)), n > 0 we get a nice series:
$$f(n)=1+\frac{ln(f(n))^1}{1!}+\frac{ln(f(n))^2}{2!}+..$$
