Nobody has yet given a combinatorial way of visualizing $e,$ so I thought I'd add one.

The constant $\pi$ is, of course, a ratio, that of the circumference of a circle to its diameter. If the corresponding ratios for regular polygons are used as a starting point, then $\pi$ is defined by taking the limit as the number of sides of the polygon goes to infinity. In a similar way, the constant $e$ is the limit of a ratio as a certain parameter is taken to infinity.

Imagine a lottery in which, to enter, you must submit a sequence of $20$ numbers in the range $1$ to $20.$ Order matters, and numbers can be used multiple times. So, for example, you can submit all $1\text{s}$ if you like. The winning number could be any of $20^{20}$ possibilities.

Now imagine that a similar lottery has been set up in a different locale, except that, because of local superstition, use of the number $13$ has been banned. Lottery entries still consist of $20$ numbers in the range $1$ to $20$, but only $19$ number choices are available, giving $19^{20}$ possible winning numbers. Clearly it's easier to win the second lottery, but by how much? The answer is the ratio $(1/19^{20})/(1/20^{20})=20^{20}/19^{20}\approx2.79.$ You are $2.79$ times as likely to win the second lottery as the first.

To generalize, imagine comparing a lottery in which submissions consist of $n$ numbers in the range $1$ to $n$ with a lottery in which one of the numbers in the range $1$ to $n$ is unlucky and may not be used. The probability of winning the second lottery is higher by a factor of $n^n/(n-1)^n.$ The limiting value of this ratio as $n$ goes to infinity is $e\approx2.718281828$.

What does this have to do with rates of growth of exponential processes? Imagine a quantity that goes up by a fixed ratio every year, say $72\%.$ Suppose we preferred to use months as units rather than years. It would be incorrect to simply divide the yearly rate by $12$ to get $72\%/12=6\%$ monthly growth, or, if we did so, we'd be describing a different process. The reason it's different is compounding. Growth of $6\%$ per month means growth by a factor of $1.06$ each month. In two months this translates to growth by a factor of $1.06\times1.06=1.1236$ or an increase of $12.36\%,$ which is bigger than $2\times6\%=12\%$ since the $6\%$ of quantity gained in the first month itself grows by $6\%$ in the second month. Similarly, over the course of a year, the factor of increase is $1.06^{12}\approx2.0122,$ which translates to $101.22\%$ growth per year instead of $72\%.$

More generally, if we prefer to use time units of $1/n$ years, an increase of $72\%/n$ every $1/n$ years is bigger than an increase of $72\%$ every year because of compounding. Specifically, it corresponds to growth by a factor of $(1+0.72/n)^n$ every year, which, if you compute it for particular $n,$ is bigger than $72\%$ per year. Moreover, if we make the time unit smaller by making $n$ bigger, the annual growth gets bigger because the compounding effect gets enhanced.

In calculus we like talking about instantaneous rates of change, which involves make the time interval over which the change is measured arbitrarily small. If we do this in our problem, by taking the limit as $n$ goes to infinity of $(1+0.72/n)^n,$ we maximize the compounding effect, obtaining a growth factor of $2.05443,$ or $105.443\%$ per year. The limiting growth factor of $2.05443$ turns out to equal to $e^{0.72}.$

To relate this to the combinatorics problem, imagine a quantity that doubles every year, that is, that increases by $100\%$ per year. If we instead split this $100\%$ over $n$ equal time intervals, we get a growth factor of $(1+1/n)^n=(n+1)^n/n^n.$ As $n$ goes to infinity, this has the same limiting value as $n^n/(n-1)^n,$ namely $e.$ Hence, because we are compounding on arbitrarily short time intervals, we grow by a factor of approximately $2.718281828$ per year rather than by a factor of $2.$

Interestingly, there's a related combinatorics problem in which $e$ appears. Imagine, once again, a lottery in which $20$ numbers in the range $1$ to $20$ must be chosen. But in this lottery all $20$ numbers must be used. So our number selection boils down to choosing a permutation of the sequence $1,2,3,\ldots,20.$ Now imagine a similar lottery with the extra stipulation that the first number may not be $1,$ the second number may not be $2,$ the third number may not be $3,$ and so on. Such a permutation is called a *derangement*. Clearly there are fewer derangements than permutations, so the second lottery is easier to win. Once again, the probability of winning the second lottery is approximately $e$ times bigger than the probability of winning the first.