Why $\lim_{n \to \infty}(1-1/n)^n = e^{-1}$? I am curious why


$$\lim_{n \to \infty} \left(1-\frac{1}{n}\right)^n = e^{-1}.$$


I can see why $$\lim_{n \to \infty} \left(1+\frac{m}{n}\right)^n = e^m$$
for some $m > 0$ by substituting $k = n/m$. But why does the result (Im guessing) also hold for the case $m <0$?
 A: Verify that for $n \ge 2$ we have
$(1- \frac{1}{n})^n=\frac{1}{(1+ \frac{1}{n-1})^{n-1}} \frac{1}{1+ \frac{1}{n-1}}$.
A: $$\left(1-\frac{1}{n}\right)^{-n}=\left(\frac{n}{n-1}\right)^{n}=\left(\frac{k+1}{k}\right)^{k}\left(\frac{k+1}{k}\right)=\left(1+\frac{1}{k}\right)^k\left(1+\frac{1}{k}\right)\to e$$
where we made the change of variable $n=k+1$
A: Another observation $$\lim_{n \to \infty}(1+1/n)^n  \times\lim_{n \to \infty}(1-1/n)^n = \\
\lim_{n \to \infty}(1+1/n)^n  (1-1/n)^n =\\
\lim_{n \to \infty}  (1-\frac{1}{n^2})^n \to 1$$ now 
we know $\lim_{n \to \infty} \left(1+\frac{1}{n}\right)^n = e^{+1}.$ 
so 
$$\lim_{n \to \infty}(1+1/n)^n  \times\lim_{n \to \infty}(1-1/n)^n =1\\
e^{+1}\times  \lim_{n \to \infty}(1-1/n)^n =1\\ \to  \lim_{n \to \infty}(1-1/n)^n =\frac{1}{e}$$
A: $$f(x)=\left(1-\frac{1}{x} \right)^x$$
Change of variable :
$$x=\frac{1}{\epsilon}\quad\to\quad f(x)=g(\epsilon)=\exp\left(\frac{\ln(1-\epsilon)}{\epsilon} \right)$$
$\frac{\ln(1-\epsilon)}{\epsilon}=\frac{1}{\epsilon}\left(-\epsilon -\frac{\epsilon^2}{2}-\frac{\epsilon^3}{3}-...\right)=-1-\frac{\epsilon}{2}-\frac{\epsilon^2}{3}-...$
$x\to\infty \qquad \epsilon \to 0 \qquad \frac{\ln(1-\epsilon)}{\epsilon} \to -1$
$$f(x\to\infty)=g(\epsilon\to 0)\to \exp(-1)$$
