II. (Full marks 50 points) Let $n \in \mathbf{N}, n \geqslant 2, k \in$ $\mathbf{R}^{+}$. Find the minimum value of the function
$$
y=\frac{x^{n}}{x-k}(x \in(k,+\infty))
$$
$$
\begin{array}{l}
\text{From the arithmetic mean-geometric mean inequality} \\
\frac{x_{1}+x_{2}+\cdots+x_{n}}{n} \geqslant \sqrt[n]{x_{1} x_{2} \cdots x_{n}}
\end{array}
$$

we get
$$
\begin{array}{l}
x_{1} x_{2} \cdots x_{n} \leqslant\left(\frac{x_{1}+x_{2}+\cdots+x_{n}}{n}\right)^{n} . \\
\because y=\frac{x^{n}}{x-b}>0 \text { and } \\
\frac{1}{y}=\frac{x-k}{x^{n}}=\frac{1}{x^{n-1}} \cdot \frac{x-k}{x}=\frac{1}{x^{n-1}}\left(1-\frac{k}{x}\right) \\
=\frac{k}{n-1}\left(\frac{1}{x}\right)\left(\frac{1}{x}\right) \cdots\left(\frac{1}{x}\right)\left(\frac{n-1}{k}-\frac{n-1}{x}\right) \\
\leqslant \frac{k}{n-1}\left(\frac{\frac{1}{x}+\cdots+\frac{1}{x}+\frac{n-1}{k}-\frac{n-1}{x}}{n}\right)^{n} \\
=\frac{k}{n-1}\left(\frac{n-1}{n k}\right)^{n}=\frac{1}{n^{n}}\left(\frac{n-1}{k}\right)^{n-1}, \\
\therefore y \geqslant n^{n}\left(\frac{k}{n-1}\right)^{n-1} .
\end{array}
$$

Equality holds if and only if $\frac{1}{x}=\frac{n-1}{k}-\frac{n-1}{x}$, i.e., $x=\frac{n k}{n-1}$.
$$
\therefore y_{\text {min }}=n^{n}\left(\frac{k}{n-1}\right)^{n-1} \text {. }
$$