which astonishingly (to me) converges quite rapidly on 1/e

To make it less surprising that (1-1/n)^n converges, here are two arguments that may help.

First: take logs. You get n log (1-1/n). Now for small x, log(1+x) = x + lower-order terms, so n log (1-1/n) = n (-1/n + lower-order terms) which obviously -> -1.

(Is it obvious enough that log(1+x) = x + lower-order terms? The easiest way to prove that might be to say that log x = integral from 1 to x of 1/t, and for x close to 1 this is roughly the integral from 1 to x of 1, or x-1.)

Second: use the binomial theorem. (1-1/n)^n = sum {k from 0 to n} of (-1)^k (n choose k) n^-k. Now (n choose k) = n(n-1)...(n-k+1) / k!, and for small k this is roughly n^k/k!. So for large n, the "early" terms are approximately +- 1/k!. And for large n, the "late" terms are relatively small because of that factor of n^-k. So (handwave handwave) you have roughly sum (-1)^k 1/k! which is the series for exp(-1).