$6 \cdot 162$ Find all functions $f:[1, \infty) \rightarrow[1, \infty)$ that satisfy the following conditions:
(1) $f(x) \leqslant 2(x+1)$;
(2) $f(x+1)=\frac{1}{x}\left[(f(x))^{2}-1\right]$.
[Proof] Clearly, the function $f(x)=x+1$ is one of the solutions we are looking for.
Next, we prove that $f(x)=x+1$ is the unique solution to this problem, i.e., we need to prove
$$
g(x)=f(x)-(x+1)=0
$$

holds for all $x$.
In fact, we can transform the conditions on $f(x)$
$$
\left\{\begin{array}{l}
1 \leqslant f(x) \leqslant 2(x+1), \\
f(x+1)=\frac{(f(x))^{2}-1}{x} .
\end{array}\right.
$$

into conditions on $g(x)$
$$
\begin{array}{l}
1 \leqslant g(x)+x+1 \leqslant 2(x+1), \\
g(x+1)+x+2=\frac{(g(x)+(x+1))^{2}-1}{x} .
\end{array}
$$

which simplifies to
$$
-x \leqslant g(x) \leqslant x+1,
$$
$$
\begin{aligned}
g(x+1) & =\frac{(g(x))^{2}+2(x+1) g(x)+(x+1)^{2}-1-x(x+2)}{x} \\
& =\frac{(g(x))^{2}+2(x+1) g(x)}{x} \\
& =\frac{g(x)[g(x)+x+x+2]}{x} .
\end{aligned}
$$

Noting that

since $g(x)+x \geqslant 0$, we have
$$
g(x)=\frac{x g(x+1)}{g(x)+x+x+2} .
$$

Thus, from $-x-1 \leqslant g(x) \leqslant x+1$, we have $|g(x)| \leqslant x+1$.

Therefore,
$$
\begin{aligned}
|g(x)| & =\frac{x|g(x+1)|}{(g(x)+x)+x+2} \leqslant \frac{x|g(x+1)|}{x+2} \\
& \leqslant \frac{x(x+2)}{x+2}=x .
\end{aligned}
$$

Thus,
$$
|g(x)| \leqslant \frac{x(x+1)}{x+2} .
$$

From this, we get $|g(x+1)| \leqslant \frac{(x+1)(x+2)}{x+3}$,
and hence $|g(x)| \leqslant \frac{\frac{x(x+1)(x+2)}{x+3}}{x+2}=\frac{x(x+1)}{x+3}$.
Let $n$ be a natural number, and $|g(x)| \leqslant \frac{x(x+1)}{x+n}$,

then $|g(x+1)| \leqslant \frac{(x+1)(x+2)}{x+n+1}$,
so
$$
\begin{aligned}
|g(x)| & \leqslant \frac{x|g(x+1)|}{x+2} \\
& \leqslant \frac{x(x+1)(x+2)}{(x+2)(x+n+1)}=\frac{x(x+1)}{x+n+1} .
\end{aligned}
$$

By the principle of mathematical induction, for all natural numbers,
$$
|g(x)| \leqslant \frac{x(x+1)}{x+n} .
$$

Taking $n \rightarrow \infty$, we conclude
$$
|g(x)| \leqslant 0, \text{ i.e., } g(x)=0 \text{ holds for all } x.
$$

This proves that
$$
f(x)=x+1
$$

is the unique solution.