$(n-1)^{(n-1)}>n^{(n-2)}$ - Chebyshev inequality 
Let $f(n)=(n-1)^{(n-1)}$ and $ g(n)=n^{(n-2)}$. Show that $f>g$ for
  all $n >2$.

I think I have to use the Binomial theorem and Chebyshev inequality to solve the problem. Is anyone is able to give me a hint?
In reality, the probelm ask to find out which number among $1999^{1999}$ and $2000^{1998}$ is the biggest one? 
 A: You could prove that $h(x)=(1+x)^{1/x}$ is decreasing for $x\geq 3$. 
Then $h(n-2)>h(n-1)$ is equivalent to your inequality.
It is actually decreasing for $x>e$ because:
$$h'(x)=h(x)\frac{1-\log x}{x^2}$$
A: Bernoulli's  inequality will do:
$$(n-1)^{n-1}>n^{n-2}\iff (n-1)\Bigl(1-\dfrac1n\Bigr)^{\!n-2}>1.$$
Now $\;\Bigl(1-\dfrac1n\Bigr)^{\!n-2}\ge1-\dfrac{n-2}n=\dfrac2n$ by Bernoulli's inequality, hence
$$ (n-1)\Bigl(1-\dfrac1n\Bigr)^{\!n-2}\ge 2-\frac2n=1+\Bigl(1-\dfrac2n\Bigr)$$
and this is $>1$ if $\;\dfrac2n<1$, i. e. if $n>2$.
A: It is equivalent to show that
$$\left(1 - \frac{1}{n}\right)^{n - 1} > \frac{1}{n}, \tag{*}$$
for $n > 2$. 
Consider a binomial random variable $X_n \sim \text{Bin}\left(n, 1 - \frac{1}{n}\right)$, then the left hand side of $(*)$ is the probability $P[X_n = n - 1]$. Since $E[X_n] = n(1 - 1/n) = n - 1 $, it then follows by Chebyshev's inequality that
\begin{align}
& \left(1 - \frac{1}{n}\right)^{n - 1} \\
= &  P[X_n - E[X_n] = 0] \\
= & P[|X_n - E[X_n]| < 1] \\
= & 1 - P[|X_n - E[X_n]| \geq 1] \\
\geq & 1 - \text{var}(X_n) \\
= & \frac{1}{n}.
\end{align}
