Why does the sum of the reciprocals of factorials converge to $e$?

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

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

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 $$ f(1) = \sum_{n=0}^\infty \frac{1}{n!}=e\,.$$