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?

If you know that $$\lim_{n \to \infty}\left(1 + \frac{1}{n}\right)^{n} = e\tag{1}$$ (and some books / authors prefer to define symbol $e$ via above equation) then it is a matter of simple algebra of limits to show that $$\lim_{n \to \infty}\left(1 - \frac{1}{n}\right)^{n} = \frac{1}{e}\tag{2}$$ Clearly we have
\begin{align}
L &= \lim_{n \to \infty}\left(1 - \frac{1}{n}\right)^{n}\notag\\
&= \lim_{n \to \infty}\left(\frac{n - 1}{n}\right)^{n}\notag\\
&= \lim_{n \to \infty}\dfrac{1}{\left(\dfrac{n}{n - 1}\right)^{n}}\notag\\
&= \lim_{n \to \infty}\dfrac{1}{\left(1 + \dfrac{1}{n - 1}\right)^{n}}\notag\\
&= \lim_{n \to \infty}\dfrac{1}{\left(1 + \dfrac{1}{n - 1}\right)^{n - 1}\cdot\dfrac{n}{n - 1}}\notag\\
&= \frac{1}{e\cdot 1}\notag\\
&= \frac{1}{e}
\end{align}
Using similar algebraic simplification it is possible to prove that $$\lim_{n \to \infty}\left(1 + \frac{x}{n}\right)^{n} = e^{x}\tag{3}$$ where $x$ is a rational number. For irrational/complex values of $x$ the relation $(3)$ holds, but it is not possible to establish it just using algebra of limits and equation $(1)$.
Regarding the intuition about "changing a tiny component in limit expression changes the output" I think it is better to visualize this simple example. We have $$\lim_{n \to \infty}n^{2}\cdot\frac{1}{n^{2}} = 1$$ and if we change the second factor $1/n^{2}$ with $(1/n^{2} + 1/n)$ then we have $$\lim_{n \to \infty}n^{2}\left(\frac{1}{n^{2}} + \frac{1}{n}\right) = \lim_{n \to \infty} 1 + n = \infty$$ The reason is very simple. The change of $1/n$ which you see here is small but due to the multiplication with other factor $n^{2}$ its impact its magnified significantly resulting in an infinite limit. You always calculate the limit of the full expression (and only when you are lucky you can evaluate the limit of a complicated expression in terms of limit of the sub-expressions via algebra of limits) and any change in a sub-expression may or may not impact the whole expression in a significant way depending upon the other parts of the expression.
