6. For $x \in \mathbf{R}, n \in \mathbf{N}_{+}$, define
$$
\mathrm{C}_{x}^{n}=\frac{x(x-1) \cdots(x-n+1)}{n!} .
$$

Let $P(x)$ be a polynomial of degree 6 that satisfies
$$
P(0)=1, P(k)=2^{k-1}(k=1,2, \cdots, 6) \text {. }
$$

Express $P(x)=$
$\qquad$

using $\mathrm{C}_{x}^{k}(k=1,2, \cdots, 6)$.
$6.1+C_{x}^{2}+C_{x}^{4}+C_{x}^{6}$.
From $P(0)=1$, we know there exists a polynomial $Q_{1}(x)$ such that
$$
P(x)=1+x Q_{1}(x) \text {. }
$$

Thus, $1=P(1)=1+Q_{1}(1)$, which implies $Q_{1}(1)=0$.
There also exists a polynomial $Q_{2}(x)$ such that
$$
Q_{1}(x)=(x-1) Q_{2}(x) \text {, }
$$

i.e., $P(x)=1+x(x-1) Q_{2}(x)$.
Thus, $2=P(2)=1+2 Q_{2}(2)$, which gives
$$
Q_{2}(2)=\frac{1}{2} \text {. }
$$

Therefore, there exists a polynomial $Q_{3}(x)$ such that
$$
Q_{2}(x)=(x-2) Q_{3}(x)+\frac{1}{2} \text {. }
$$

Then, $P(x)=1+\mathrm{C}_{x}^{2}+x(x-1)(x-2) Q_{3}(x)$.
From $4=P(3)=1+3+3!Q_{3}(3)$, we know
$$
Q_{3}(3)=0 \text {. }
$$

Thus, $Q_{3}(x)=(x-3) Q_{4}(x)$,
$$
P(x)=1+\mathrm{C}_{x}^{2}+x(x-1)(x-2)(x-3) Q_{4}(x) \text {. }
$$

Further, we have
$$
\begin{aligned}
P(x)= & 1+C_{x}^{2}+C_{x}^{4}+ \\
& x(x-1)(x-2)(x-3)(x-4) Q_{5}(x) .
\end{aligned}
$$

Continuing this process and using the fact that $P(x)$ is a 6th-degree polynomial, we get
$$
P(x)=1+C_{x}^{2}+C_{x}^{4}+C_{x}^{6} .
$$