How to prove this inequality about $e$? 
Possible Duplicate:
Proving $(1 + 1/n)^{n+1} \gt e$ 

How to prove this:
$$
\left(\frac{x}{x-1}\right)^x \geq e
\qquad\text{for}\qquad
x \in \mathbb{N}^*
$$
$e$ is the base of the natural logarithm.
and I think the equal satisfies if $x$ converges to infinity.
Thank you!

First off, $\frac{x}{x-1} > 0$ iff $x < 0$ or $x > 1$,
so we can't take the natural logarithm if $x\in[0,1]$.
My answer addresses the inequality for real-valued $x$, as in the original post
(proving it for $x > 1$ and disproving it for $x < 0$).
Now
$$
e\le
\left(\frac{x}{x-1}\right)^{x}=
\left(1-\frac1x\right)^{-x}
\to e
$$
already guarantees the limiting behavior for us.
Depending on the sign of $x$, this inequality becomes
$$
e^{-1/x} \le 1-\frac1x \qquad(x < 0)
$$
$$
e^{-1/x} \ge 1-\frac1x \qquad(x > 1)
$$
Setting $t=\frac1x\lt1$ and using the Taylor series,
this translates to
$$
\eqalign{
1-t &\le e^{-t} = \sum_{n=0}^\infty\frac{(-1)^n t^n}{n!} \\
    &\le 1-t+\frac{t^2}2+\frac{t^3}6+\cdots
}
$$
for $t \in (0,1)$, which is patently true,
while for negative values of $t$,
the reversed inequality is patently false
(which we can easily check in the original by trying $x=-1$ since $2 < e$).
Therefore I would suggest adding the stipulation that
$x > 0$ (necessarily) or actually, $x > 1$.
