Eafe ** Let $n$-degree polynomial $P(x)$ satisfy: for $k=1,2,4, \cdots, 2^{n}$, we have $P(k)=\frac{1}{k}$. Find $P(0)$.
From the conditions, we know that $1,2,2^{2}, \cdots, 2^{n}$ are all roots of the polynomial $x P(x)-1$. Since $P(x)$ is an $n$-degree polynomial, we can assume
$$
x P(x)-1=a(x-1)(x-2) \cdots\left(x-2^{n}\right),
$$

where $a$ is the leading coefficient of $P(x)$.
Substituting $x=0$ into (1), we get
$$
-1=a \cdot(-1)(-2) \cdots\left(-2^{n}\right)=(-1)^{n+1} \cdot a \cdot 2^{\frac{n(n+1)}{2}},
$$

Thus, $a=(-1)^{n} \cdot 2^{-\frac{n(n+1)}{2}}$. Now, comparing the coefficients of $x$ on both sides of (1), we find
$$
\begin{aligned}
P(0) & =a \cdot 2^{\frac{n(n+1)}{2}} \cdot(-1)^{n}\left(1+\frac{1}{2}+\frac{1}{2^{2}}+\cdots+\frac{1}{2^{n}}\right) \\
& =1+\frac{1}{2}+\frac{1}{2^{2}}+\cdots+\frac{1}{2^{n}}=2-\frac{1}{2^{n}} .
\end{aligned}
$$

Therefore, $P(0)=2-\frac{1}{2^{n}}$.