Prove that $\lim_{x \to \infty}\big(\frac{x}{x-1}\big)^x$ is also $e$. Trying to make sense out of the idea that $100\%$ continuous decay is $\frac{1}{e}$, I thought about this:
You can express $1+\frac{1}{x}$ as $\frac{x+1}{x}$, such that $\big(1+\frac{1}{x}\big)^x = \big(\frac{x+1}{x}\big)^x$
And you can express $1-\frac{1}{x}$ as $\frac{x-1}{x}$, such that $\big(1-\frac{1}{x}\big)^x = \big(\frac{x-1}{x}\big)^x =\frac{1}{\big(\frac{x}{x-1}\big)^x}$
Now $\big(\frac{x}{x-1}\big)^x$ and $\big(\frac{x+1}{x}\big)^x$ look very similar, and I can imagine (and see on Mathematica) that $\lim_{x \to \infty} \big(\frac{x}{x-1}\big)^x$ also approaches $e$.
However I'm clueless about limit proofs, how do you prove that?
EDIT: Just had a last second insight.  Can you say that $$\lim_{x \to \infty} \big(\frac{x}{x-1}\big)^x = \lim_{x \to \infty} \big(\frac{x}{x-1}\big)^{x-1} \cdot \lim_{x \to \infty} \big(\frac{x}{x-1}\big)$$ $$= \lim_{x \to \infty} \big(\frac{x}{x-1}\big)^{x-1} \cdot 1$$ $$=\lim_{x \to \infty} \big(\frac{x+1}{x}\big)^x = e ?$$ 
(not sure about that last limit transition...I mean I'm pretty sure it's true, just not sure how to write it; I would appreciate feedback on that, thank you)
 A: $$
\lim_{n\to \infty} \left(\frac{n}{n-1}\right)^n=\lim_{n\to \infty} \left(1+\frac{1}{n-1}\right)^{n-1} \cdot \left(1+\frac{1}{n-1}\right)= e.
$$
A: For your second insight...
$$\lim\limits_{x \to \infty}\left(\dfrac{x}{x-1}\right)^{x}$$
You can split the limit into product terms if and only if the product terms both exist.
i.e., before splitting into a product, observe that
$$\lim\limits_{x \to \infty}\left(\dfrac{x}{x-1}\right)^{x-1} = \lim\limits_{t \to \infty}\left(\dfrac{t+1}{t}\right)^{t}$$
using the substitution $t = x - 1$ (and as $x \to \infty$, intuitively, $t \to \infty$ as well). Obviously, this is equal to $e$.
Also observe that $\lim\limits_{x \to \infty}\left(\dfrac{x}{x-1}\right) = 1$.
Because the product limits exist, 
$$\lim\limits_{x \to \infty}\left(\dfrac{x}{x-1}\right)^{x} = \left[ \lim\limits_{x \to \infty}\left(\dfrac{x}{x-1}\right)^{x-1}\right]\left[ \lim\limits_{x \to \infty}\left(\dfrac{x}{x-1}\right)\right] = e \cdot 1 = e\text{.}$$
A: $$\left(\frac{x}{x-1}\right)^x=\left(\frac{x-1+1}{x-1}\right)^x=\left(1+\frac{1}{x-1}\right)^x=\left[\left(1+\frac{1}{x-1}\right)^{x-1}\right]^{\frac{x}{x-1}}$$
               When $x\to \infty$, the exponent $\frac{x}{x-1}=\frac{1}{1-1/x}$ tends to $1$ and $\left(1+\frac{1}{x-1}\right)^{x-1}$ it is known tends to $e$.
A: If you know that
$$
\lim_{x\to\infty}\left(1+\frac{a}{x}\right)^{x}=e^a
$$
then
$$
\lim_{x\to\infty}\left(1-\frac{1}{x}\right)^{x}=e^{-1}
$$
and so
$$
\lim_{x\to\infty}\left(\frac{x}{1-x}\right)^x=
\lim_{x\to\infty}\frac{1}{\left(\dfrac{x-1}{x}\right)^x}=
\lim_{x\to\infty}\frac{1}{\left(1-\dfrac{1}{x}\right)^x}=
(e^{-1})^{-1}=e
$$
The proof of the first limit is easy:
$$
\lim_{x\to\infty}\log\left(1+\frac{a}{x}\right)^{x}=
\lim_{x\to\infty}x\log\left(1+\frac{a}{x}\right)=
\lim_{t\to0^+}\frac{\log(1+at)}{t}=a
$$
because it's the derivative at $0$ of the function $t\mapsto\log(1+at)$.
