6. Given the sequence $\left\{a_{n}\right\}$ satisfies
$$
\begin{array}{l}
a_{0}=0, \\
a_{n}=a_{n-1}^{2}+2 n a_{n-1}+n^{2}-n-1\left(n \in \mathbf{N}_{+}\right) .
\end{array}
$$

Then $S_{n}=\sum_{k=1}^{n} a_{k} \mathrm{C}_{n}^{k}=$ $\qquad$
6. $-2^{n-1} n$.

From the recurrence relation, we get
$$
\begin{array}{l}
a_{n}+(n+1)=\left(a_{n-1}+n\right)^{2} \\
=\left[a_{n-2}+(n-1)\right]^{2 \times 2}=\cdots \\
=\left(a_{0}+1\right)^{2 n}=1
\end{array}
$$

Thus, $a_{n}=-n$.
Then, $S_{n}=-\sum_{k=1}^{n} k \mathrm{C}_{n}^{k}=-\sum_{k=1}^{n} n \mathrm{C}_{n-1}^{k-1}=-2^{n-1} n$.