Example 11 If $p(x)$ is an $n$-degree polynomial, and for $k=0,1,2, \cdots, n$, $p(k)=\frac{k}{k+1}$. Try to determine $p(n+1)$.
(4th USA Mathematical Olympiad)
Let $Q(x) = (x+1) \cdot p(x) - x$. From the given conditions, we know that $Q(x)$ is a polynomial of degree no more than $n+1$, and when $x=0,1,2, \cdots, n$, we have $Q(x)=0$.
Thus, we have $Q(x) = a x(x-1)(x-2) \cdots(x-n)$, where $a$ is a constant, i.e., $(x+1) p(x) - x = a x(x-1)(x-2) \cdots(x-n)$.
Let $x=-1$, we get $1 = a(-1)^{n+1} \cdot (n+1)!$,
which means $a = \frac{(-1)^{n+1}}{(n+1)!}$.
Therefore, $p(x) = \frac{Q(x) + x}{x+1} = \frac{\frac{(-1)^{n+1}}{(n+1)!}(x-1)(x-2) \cdots(x-n) + x}{x+1}$.
Then $p(n+1) = \frac{(-1)^{n+1} + n + 1}{n + 2}$.
Hence, $p(n+1) = \left\{\begin{array}{l}1, \text{ if } n \text{ is odd; } \\ \frac{n}{n+2}, \text{ if } n \text{ is even. }\end{array}\right.$