Three. (20 points) Given the function
$$
f(x)=x^{n}-x^{n+1}\left(x \in[0,1], n \in \mathbf{N}_{+}\right) \text {. }
$$
(1) Find the maximum value $g(n)$ of $f(x)$;
(2) Find the maximum value of $g(n)$.
$$
\begin{array}{l}
\text { (3) From } f^{\prime}(x)=n x^{n-1}-(n+1) x^{n} \\
=x^{n-1}[n-(n+1) x],
\end{array}
$$

we can see that $f(x)$ is monotonically increasing on $\left[0, \frac{n}{n+1}\right]$ and monotonically decreasing on $\left[\frac{n}{n+1}, 1\right]$.
Therefore, at $x=\frac{n}{n+1}$, $f(x)$ reaches its maximum value
$$
g(n)=\frac{n^{n}}{(n+1)^{n+1}} \text {. }
$$
(2) $g(1)=\frac{1}{4}$.

Next, we prove: When $n \geqslant 2$, $g(n)4 n^{n}$.
$$
\begin{array}{l}
\text { From }(n+1)^{n+1}=n^{n+1}+\mathrm{C}_{n+1}^{1} n^{n}+\cdots+\mathrm{C}_{n+1}^{n+1} \\
\geqslant n^{n+1}+(n+1) n^{n}=(2 n+1) n^{n}>4 n^{n},
\end{array}
$$

we know that $g(n)<\frac{1}{4}$, i.e., the maximum value of $g(n)$ is $\frac{1}{4}$.