Intuitive explanation for why $\left(1-\frac1n\right)^n \to \frac1e$ I am aware that $e$, the base of natural logarithms, can be defined as:
$$e = \lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n$$
Recently, I found out that
$$\lim_{n\to\infty}\left(1-\frac{1}{n}\right)^n = e^{-1}$$
How does that work? Surely the minus sign makes no difference, as when $n$ is large, $\frac{1}{n}$ is very small?
I'm not asking for just any rigorous method of proving this. I've been told one: as $n$ goes to infinity, $\left(1+\frac{1}{n}\right)^n\left(1-\frac{1}{n}\right)^n = 1$, so the latter limit must be the reciprocal of $e$. However, I still don't understand why changing such a tiny component of the limit changes the output so drastically. Does anyone have a remotely intuitive explanation of this concept?

Actually you have the stronger true statement that
$$
  \lim_{x\to0}(1+x)^{1/x}=e,
$$
of which the initial limit you stated is a special case, approaching $0$ through the sequence of values $x=\frac1n$ for $n\in\Bbb N_{>0}$. But if you approach $0$ through the sequence of values $x=-\frac1n$ for $n\in\Bbb N_{>1}$, the same limit gives you
$$
  \lim_{n\to\infty}\left(1-\frac1n\right)^{-n}=e.
$$
Now it is a simple matter to see that the sequence of inverses $\left(1-\frac1n\right)^n$ tends to the inverse value $e^{-1}$.
It should be noted that while the first limit above is more general than the limits for $n\to\infty$, it is also less elementary to define, since it involves powers of positive real numbers with arbitrary real exponents. Introducing such powers requires studying exponential functions in the first place, which is why the limit statement with integer exponents is often preferred. But the more general limit statement is true, and can serve to give intuition for the relation between the two limits in your question.
