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.

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, $$\left(1+\frac1{1000}\right)^{1000}=\frac1{0!}+\frac1{1!}+\frac{0.999}{2!}+\frac{0.997002}{3!}+\frac{0.994010994}{4!}\dots$$

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

$$ e = \sum_{n=0}^\infty \frac{1}{n!}, $$

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 $e$).