22. (12 points) Given the function
$$
f(x)=a x-1-\ln x(a \in \mathbf{R}) \text {. }
$$
(1) Discuss the number of extremum points of the function $f(x)$ within its domain;
(2) If the function $f(x)$ has an extremum at $x=1$, and for any $x \in(0,+\infty), f(x) \geqslant b x-2$ always holds, find the range of the real number $b$;
(3) When $x>y>\mathrm{e}-1$, prove:
$$
\mathrm{e}^{x-y}>\frac{\ln (x+1)}{\ln (y+1)} .
$$
22. (1) Note that, the domain of the function $f(x)$ is $(0,+\infty), f^{\prime}(x)=a-\frac{1}{x}$.

When $a \leqslant 0$, $f^{\prime}(x) < 0$, so $f(x)$ is monotonically decreasing on $(0,+\infty)$.

When $a > 0$, $f(x)$ is monotonically decreasing on the interval $\left(0, \frac{1}{a}\right)$ and monotonically increasing on the interval $\left(\frac{1}{a},+\infty\right)$. Therefore, the function $f(x)$ has one extremum point.
(2) Given that the function $f(x)$ attains an extremum at $x=1$, we have $a=1$.
Thus, $f(x) \geqslant b x-2 \Leftrightarrow 1+\frac{1}{x}-\frac{\ln x}{x} \geqslant b$. Let $g(x)=1+\frac{1}{x}-\frac{\ln x}{x}$, then the function $g(x)$ is monotonically decreasing on the interval $\left(0, \mathrm{e}^{2}\right]$ and monotonically increasing on the interval $\left[\mathrm{e}^{2},+\infty\right)$.
Hence, $g(x)_{\min }=g\left(\mathrm{e}^{2}\right)=1-\frac{1}{\mathrm{e}^{2}}$, which means $b \leqslant 1-\frac{1}{\mathrm{e}^{2}}$.
(3) Note that,
$\mathrm{e}^{x-y}>\frac{\ln (x+1)}{\ln (y+1)} \Leftrightarrow \frac{\mathrm{e}^{x}}{\ln (x+1)}>\frac{\mathrm{e}^{y}}{\ln (y+1)}$.
Let $m(x)=\frac{\mathrm{e}^{x}}{\ln (x+1)}$.
Then $m^{\prime}(x)=\frac{\mathrm{e}^{x}\left(\ln (x+1)-\frac{1}{x+1}\right)}{\ln ^{2}(x+1)}$.
Clearly, the function $h(x)=\ln (x+1)-\frac{1}{x+1}$ is monotonically increasing on the interval $(\mathrm{e}-1,+\infty)$.
Thus, $h(x)>1-\frac{1}{\mathrm{e}}>0$, which means $m^{\prime}(x)>0$.
Therefore, $m(x)$ is monotonically increasing on the interval $(\mathrm{e}-1,+\infty)$, i.e.,
$$
\frac{\mathrm{e}^{x}}{\ln (x+1)}>\frac{\mathrm{e}^{y}}{\ln (y+1)} \Rightarrow \mathrm{e}^{x-y}>\frac{\ln (x+1)}{\ln (y+1)} .
$$