7. For $x \in \mathbf{R}, n \in \mathbf{N}^{*}$, define $C_{x}^{n}=\frac{x(x-1) \cdots(x-n+1)}{n!}$. Let $P(x)$ be a 6th degree polynomial that satisfies $P(0)=1, P(k)=2^{k-1}, k=1,2, \cdots, 6$, express $P(x)=$ using $C_{x}^{k}(k=1,2, \cdots, 6)$.
7. $P(x)=1+C_{x}^{2}+C_{x}^{4}+C_{x}^{6}$ Detailed Explanation: From $P(0)=1$ we know there exists a polynomial $Q_{1}(x)$ such that $P(x)=1+x Q_{1}(x)$, hence $1=P(1)=1+Q_{1}(1)$, so $Q_{1}(1)=0$. There also exists a polynomial $Q_{2}(x)$ such that $Q_{1}(x)=(x-1) Q_{2}(x)$, i.e., $P(x)=1+x(x-1) Q_{2}(x)$. Therefore, $2=P(2)=1+2 Q_{2}(2)$, so $Q_{2}(2)=\frac{1}{2}$. Thus, there exists a polynomial $Q_{3}(x)$ such that $Q_{2}(x)=(x-2) Q_{3}(x)+\frac{1}{2}$, hence
$P(x)=1+C_{x}^{2}+x(x-1)(x-2) Q_{3}(x)$. Also, $4=P(3)=1+3+3! Q_{3}(3)$, so $Q_{3}(3)=0$, hence $Q_{3}(x)=(x-3) Q_{1}(x), P(x)=1+C_{x}^{2}+x(x-1)(x-2)(x-3) Q_{1}(x)$. Further, $P(x)=1+C_{x}^{2}+C_{x}^{4}+x(x-1)(x-2)(x-3)(x-4) Q_{5}(x)$.
Continuing this process and using the fact that $P(x)$ is a 6th degree polynomial, we get $P(x)=1+C_{x}^{2}+C_{x}^{4}+C_{x}^{6}$.