Proof: $\lim_{n\to\infty} (1-\frac{1}{n})^{-n} = e$ If $e:=\lim_{n\to\infty} (1 + \frac{1}{n})^n$, prove that $$ \lim_{n\to\infty} \left(1-\frac{1}{n}\right)^{-n} = e $$
  without using the property that says: if $\lim_{n\to\infty} a_n = \infty$, $\lim_{n\to\infty} x_n = x$, then $\lim{n\to\infty}(1+\frac{x_n}{a_n})^{a_n} = e^x$...


I've tried rewriting $1-\frac{1}{n}$, and operating $\lim_{n\to\infty}(1-\frac{1}{n})^n = 1 / \lim_{n\to\infty}(1+\frac{1}{n})^n $ but I couldn't prove it. Any hint? Thanks in advance.
\begin{gather}
\lim\limits_{n\to\infty}\left(1-\dfrac{1}{n}\right)^{-n}=\lim\limits_{n\to\infty}\left(\dfrac{n-1}{n}\right)^{-n}= \\
=\lim\limits_{n\to\infty}\left(\dfrac{n}{n-1}\right)^{n}=\lim\limits_{n\to\infty}\left[\left(1+\dfrac{1}{n-1}\right)^{n-1}\left(1+\dfrac{1}{n-1}\right)\right]=e
\end{gather}
To prove that $(1-\frac{1}{x})^{-x} = e$, as x goes to infinity, you would have to prove that $ln(1-\frac{1}{x})^{-x}=1$ under the same conditions.I have taken natural log of both sides.

I will now use y=1/x for convenience.
In other words,I'd have to prove that when x=1/y, y going to zero:

$(1/-y)*ln(1-y) =(ln(1-y))/(-y)=1$

We have an indefinite form here.Infinity in the numerator, 0 in the denominator.so we can use L'Hopitals Rule .
We take the derivatives of both numerator and denominator:

$(-1/(1-y))/-1)$, and this gives 1/1 when y goes to zero.In other words, when x ,being 1/y, goes to infinity , $ln(1-\frac{1}{x})^{-x}=1$ in the limit, which had to be proven.