8. Let $P_{n}(k)$ denote the number of permutations of $\{1,2, \cdots, n\}$ with $k$ fixed points. Let $a_{t}=\sum_{k=0}^{n} k^{t} P_{n}(k)$. Then
$$
\begin{array}{l}
a_{5}-10 a_{4}+35 a_{3}-50 a_{2}+25 a_{1}-2 a_{0} \\
=
\end{array}
$$
8. 0 .

Notice
$$
\begin{array}{l}
a_{5}-10 a_{4}+35 a_{3}-50 a_{2}+25 a_{1} \\
=\sum_{k=0}^{n}\left(k^{5}-10 k^{4}+35 k^{3}-50 k^{2}+25 k\right) P_{n}(k) \\
=\sum_{k=0}^{n}[k(k-1)(k-2)(k-3)(k-4)+k] P_{n}(k) \\
=\sum_{k=0}^{n}[k(k-1)(k-2)(k-3)(k-4)+k] . \\
\frac{n!}{(n-k)!k!} P_{n-k}(0) \\
=\sum_{k=0}^{n}\left[\frac{n!}{(n-5)!} \cdot \frac{(n-5)!}{(k-5)!(n-k)!} P_{(n-5)-(k-5)}(0)+\right. \\
\left.n \cdot \frac{(n-1)!}{(k-1)!(n-k)!} P_{(n-1)-(k-1)}(0)\right] \\
=\frac{n!}{(n-5)!} \sum_{k=0}^{n-5} P_{n-5}(k)+n \sum_{k=0}^{n-1} P_{n-1}(k) .
\end{array}
$$

Since $a_{0}=\sum_{k=0}^{n} P_{n}(k)=n!$, therefore,
$$
\begin{array}{l}
a_{5}-10 a_{4}+35 a_{3}-50 a_{2}+25 a_{1}-2 a_{0} \\
=\frac{n!}{(n-5)!}(n-5)!+n \cdot(n-1)!-2 \cdot n! \\
=0 .
\end{array}
$$