$k$ fixed points. We denote with $p_n(k)$ the number of permutations of the numbers $1,2,...,n$ that have exactly $k$ fixed points.
a) Prove that $\sum_{k=0}^n kp_n (k)=n!$.
b) If $s$ is an arbitrary natural number, then:
$\sum_{k=0}^n k^s p_n (k)=n!\sum_{i=1}^m R(s,i)$,
where with $R(s,i)$ we denote the number of partitions of the set $\{1,2,...,s\}$ into $i$ non-empty non-intersecting subsets and $m=min(s,n)$.
Attemp: 
a) $$\sum_{k=1}^{n}kp_n(k)$$$$=\sum_{k=1}^{n}k \binom{n}{k} D_{n-k}$$$$= \sum_{k=1}^{n} \left (k \frac{n!}{k!} (n-k)! \sum_{i=1}^{n-k} \frac{(-1)^{i+1}}{i!} \right)$$$$=\sum_{k=1}^{n} \frac{n! \sum_{i=1}^{n-k} \frac{(-1)^{i+1}}{i!}}{(k-1)!}$$$$=n! \sum_{k=1}^{n} \frac{1}{e(k-1)!}=\frac{n!}{e} \sum_{k=1}^{n} \frac{1}{(k-1)!}$$$$=n!$$
No ideas for b)
 A: Your solution has several missteps. In your third equality, an $(n-k)!$ disappeared. Also, the equation $\sum_{i=1}^{n-k}\frac{(-1)^{i+1}}{i!}= e^{-1}$ used in your fourth equality is incorrect.

Here is a combinatorial solution. $\sum_{k=1}^nkp_k(n)$ counts the number of ordered pairs $(\pi,j)$, where $\pi$ is a permutation and $j$ is a fixed point of $\pi$. This is because there are $p_k(n)$ permutations with $k$ fixed points, and for each of those, there are $k$ choices for $j$. 
To count the number of ordered pairs $(\pi,j)$, let us ask how many choices of $\pi$ there are for a given $j$. The number of permutations which have a given number $j$ as a fixed point is $(n-1)!$. Therefore,
$$
\begin{align}
\sum_{k=1}^nkp_k(n)
&=\text{# of $(\pi,j)$ such that $j$ is fixed point of $\pi$}
\\&=\sum_{j=1}^n \text{# of $\pi$ such that $j$ is fixed point of $\pi$}
\\&=\sum_{j=1}^n (n-1)!
\\&= n\cdot (n-1)!=n!.
\end{align}
$$
For $(b)$, you are instead counting ordered sequences $(\pi,j_1,j_2,\dots,j_s)$ so that $j_1,\dots,j_s$ are all fixed points of $\pi$, as there are $k^s$ ways to choose the list of $j_1,\dots,j_s$ when $\pi$ has $k$ fixed points. Given a list which contains $i$ distinct values, there are $(n-i)!$ ways to choose a permutation $\pi$ which has all those values as fixed points. 
The number of lists $j_1,\dots,j_s$ which have exactly $i$ distinct values is $$R(s,i)\cdot \frac{n!}{(n-i)!}.$$This is because such a list is formed by partitioning $\{1,2,\dots,s\}$ into $n$ parts, ordering the parts by smallest element, then assigning the first part one of $n$ numbers, then assigning the second one one of $(n-1)$ numbers different than the first, and so on.
Therefore,
$$
\begin{align}
\sum_{k=1}^nk^sp_k(n)
&=\text{# of $(\pi,j_1,\dots,j_s)$ such that each $j_i$ is fixed point of $\pi$}
\\&=\sum_{i=1}^s(\text{# $(j_1,\dots,j_s)$ w/ $i$ distinct values})\cdot(\text{# of $\pi$ where $i$ particular points are fixed})
\\&=\sum_{i=1}^s R(s,i)\cdot \frac{n!}{(n-i)!}\cdot (n-i)!
\\&= n!\sum_{i=1}^s R(s,i).
\end{align}
$$
