Limit for $e$ and $\frac{1}{e}$ My question concerns the derivation of this: $$e^r = \lim_{n \rightarrow \infty} \left(1 + \frac{r}{n}\right)^n \ \ ...(1).$$
One of the definitions of $e$ is as follows:
$$e = \lim_{n \rightarrow \infty} \left(1+\frac{1}{n}\right)^n.$$
Then, textbooks usually derive equation (1) in the following manner:
\begin{align}
\lim_{n \rightarrow \infty} \left(1 + \frac{r}{n}\right)^n &= \lim_{u \rightarrow \infty} \left(1 + \frac{1}{u}\right)^{ru} \ \ \text{where} \ u = \frac{n}{r}\\
&= \lim_{u \rightarrow \infty} \left(\left(1 + \frac{1}{u}\right)^u\right)^r \\
&= \left(\lim_{u \rightarrow \infty} \left(1 + \frac{1}{u}\right)^u\right)^r \\
&= e^r.
\end{align}
This argument is fine if $r > 0$ since $u \rightarrow \infty$ as $n \rightarrow \infty$, but when $r < 0$, $u \rightarrow - \infty$ as $n \rightarrow \infty$.
How can I extend the proof for (1) where $r$ is any real number?
When $r = 0$, $\lim_{n \rightarrow \infty} (1+0/n)^n = 1$ (Although, I should be careful about evaluating limits that look like $``1^{\infty}"$.)
Here's my attempt so far for the case where $r < 0$:
\begin{align}
\lim_{n \rightarrow \infty} \left(1 + \frac{r}{n}\right)^n &= \lim_{u \rightarrow \infty} \left(1 - \frac{1}{u}\right)^{-ru} \ \text{where} \ u = -\frac{n}{r} \\
&= \left(\lim_{u \rightarrow \infty} \left(1 - \frac{1}{u}\right)^u\right)^{-r}.
\end{align}
My question boils down to how to show the following limit from the definition above for $e$ $$\lim_{n \rightarrow \infty} \left(1 - \frac{1}{n}\right)^n = \frac{1}{e}.$$
Thanks.

Consider proving the reciprocal, i.e.
$$
\lim \left(1 - \frac 1n\right)^{-n} = \mathrm e. 
$$
The expression inside could be rewritten as 
$$
\left(1 - \frac 1n \right)^{-n} = \left(\frac {n-1}n\right)^{-n} = \left(\frac n{n-1}\right)^n = \left(1 + \frac 1{n-1}\right)^n = \left(1 + \frac 1{n-1}\right)^{n-1} \cdot \left(1 + \frac 1{n-1}\right). 
$$
Now use the definition of $\mathrm e$:
$$
\lim_n \left(1 + \frac 1{n-1}\right)^{n-1} = \mathrm e.
$$
Since 
$$
\lim_n \left(1 + \frac 1{n-1}\right) = 1, 
$$
we conclude that 
$$
\lim_n \left(1 - \frac 1n \right)^{-n} = \lim_n \left(1 + \frac 1{n-1}\right)^{n-1} \cdot \lim_n \left(1 + \frac 1{n-1}\right) = \mathrm e \cdot 1 = \mathrm e. 
$$
by the law of arithmetic operations of limits. 
Therefore the original limit is $1/\mathrm  e$.
