Showing that $\lim_{n \rightarrow \infty}\left(\frac{n - 1}{n}\right)^n = 1/e$ I would like to show  $\lim_{n \rightarrow \infty}\left(\frac{n - 1}{n}\right)^n = 1/e$.
I know the argument typically goes like this: Let $y = \left(\frac{n - 1}{n}\right)^n$. Then $\ln(y) = n\cdot{}\ln \left(\frac{n - 1}{n}\right)$. Taking the limit as $n \rightarrow \infty$, we have an indeterminant product of the form $\infty\cdot0$.
I think ideally I would like to use L'Hopital's Rule, so the issue is getting this into the correct form to apply it. I don't think the simplification $n\ln(n - 1) - n\ln(n)$ helps any.
But if we can establish that $\lim_{n\rightarrow\infty}\ln(y) = -1$, then using the identity $y = e^{\ln(y)}$, we'd arrive at the desired result.
Alternatively, could one use the definition of $e$? This might not help, but $e = \lim_{n \rightarrow \infty}(1 + \frac{1}{n})^n = \lim_{n \rightarrow \infty}(\frac{n + 1}{n})^n$, which looks similar to what we have, but not quite.
 A: The idea is good; we try proving that
$$
\lim_{x\to\infty}\left(\frac{x-1}{x}\right)^x
$$
exists; if it does, then it's the same as the limit of your sequence. So we try finding the limit of the logarithm:
$$
\lim_{x\to\infty}x\log\frac{x-1}{x}
$$
Now do the substitution $x=1/t$, which brings the limit to the form
$$
\lim_{t\to0^+}\frac{1}{t}\log(1-t)
$$
which is the derivative at $0$ of the function $f(t)=\log(1-t)$.
Why doing $t=1/x$? For two reasons. First
$$
\frac{x-1}{x}=1-\frac{1}{x}=1-t
$$
Second, we have $x$ as a factor, which becomes $t$ at the denominator. Third, this transforms a limit at $\infty$ to a limit at $0$, where derivatives are defined.
A: You can use the definition $e = \lim_{n \rightarrow \infty} (1+1/n)^n$.
Note that
$$\lim\left(\frac{n-1}{n}\right)^{-n}= \lim\left[\left(1+\frac{1}{n-1}\right)^{n-1}\left(1+\frac{1}{n-1}\right)\right]\\= \lim\left(1+\frac{1}{n-1}\right)^{n-1}\lim\left(1+\frac{1}{n-1}\right)= e$$
Hence,
$$\lim\left(\frac{n-1}{n}\right)^{n}= e^{-1}$$
A: This is a definition of Euler's number as well, 
I have found a formula that is asymptotic about a week ago while playing around with such limits that is:
$$
\lim_{n \rightarrow \infty} ((n+1)!^{\frac{1}{n+1}} - (n)!^{\frac{1}{n}}) \sim \lim_{n \rightarrow \infty} (1-\frac{1}{n})^{n}
$$ 
where Traian Lalescu has proven that  this is true in a rigorous proof which is a more fun way of proofing but takes alot more work
also you can simplify $\frac{n-1}{n}$ = $\frac{n}{n+1}$ which you can use the known limit
$$
\lim_{n \rightarrow \infty} (\frac{n+1}{n})^n = e
$$
link:
http://mathhelpforum.com/differential-geometry/102384-solved-prove-lim-1-1-n-n-1-e.html
A: Consider making the change of variable $k=n-1$ here
to get $$\lim_{n\rightarrow\infty}\left(\frac{n-1}{n}\right)^n = \lim_{k\rightarrow\infty}\left(\frac{k}{k+1}\right)^{k+1}
=\lim_{k\rightarrow\infty}\left(\frac{k}{k+1}\right)\lim_{k\rightarrow\infty}\left(\frac{k}{k+1}\right)^{k}
=\lim_{k\rightarrow\infty}\left(\frac{k}{k+1}\right)^{k}$$
which looks like the reciprocal of the limit definition of $e$.
If you want to use your original approach you can write $\ln(y)$ as
$$\lim_{n\rightarrow\infty}\frac{\ln(\frac{n-1}{n})}{\frac{1}{n}}$$ which has indeterminate form $\frac{0}{0}$ so we use L'Hopital's to get this as 
$$\lim_{n\rightarrow\infty}\frac{\frac{n}{n-1}\frac{1}{n^2}}{\frac{-1}{n^2}} =
-\lim_{n\rightarrow\infty}\frac{n}{n-1 } = -1 $$
A: You can even do
$$ \lim \left(\frac{n-1}{n}\right)^{n} = \lim \left(\frac{1}{\frac{n}{n-1}}\right)^{n} = \lim \left({\frac{1}{\frac{n+1}{n}}}\right)^{n}$$
With the last equality holding because we are taking the limit as $n$ approaches infinity. Thus, we have
$$\lim\left(\frac{1^{n}}{e}\right) = \frac{1}{e}$$ as desired.
A: We have that
$$(\frac{n-1}{n})^n=(1-\frac{1}{n})^n=(1+\frac{-1}{n})^n$$
We have the classical limit
$$\lim_{x\to\infty}(1+\frac{k}{n})^n=e^k$$
our $k=-1$ so
$$\lim_{x\to\infty}(1+\frac{-1}{n})^n=e^{-1}=\frac{1}{e}$$
