Prove $C_n=B_{n-1}-B_{n-2}+\dots +(-1)^nB_1$ If $C_n$ is the number of the ways to partition set $\{1,2,\dots ,n \}$ into subsets that has at least two members and $B_n$ is the number of the ways to partition the same set into non empty subsets then prove that:
$C_n=B_{n-1}-B_{n-2}+\dots +(-1)^nB_1$
It seems to we have to use inclusion-exclution principle But how can we connect $C_n$ to $B_{n-1}$?First I thought we should take a member and partition the others to some non-empty sets but it didn't work.Any hints?
 A: The following argument is based on the idea that 
every partition of $\{1,\cdots, n\}$ with a singleton 
corresponds to a partition of $\{1,\cdots, n+1\}$ without a singleton.

Let $T_n$ be the set of partitions of $[n]$ with at least 2 elements in each subset of the partition, and
let $P=\{A_1,\cdots,A_m\}$ be in $Q$.
1)$\;\;$ If $|A_i|\ge2$ for $1\le i\le m$, then $P\in T_n$.
2)$\;\;$ If $P\not\in T_{n}, \text{ with }|A_i|\ge2$ for $1\le i\le r$ and $A_i=\{a_i\}$ for $r+1\le i\le m$ (where $r<m$),  
$\hspace{.27 in}$let $P^{\prime}=\{A_1,\cdots,A_r,B\}$ where $B=\{a_{r+1},\cdots,a_m,n+1\};\;$ so $P^{\prime}\in T_{n+1}$.
This sets up a 1-1 correspondence between partitions of $[n]$ which are not in $T_n$ and partitions of $[n+1]$ which are in $T_{n+1}$, since $P^{\prime}\in T_{n+1}$ with $P^\prime=\{B_1,\cdots,B_r, B_{r+1}\}$ 
and $B_{r+1}=\{a_1,\cdots,a_l,n+1\}$ corresponds to the partition $P=\{B_1,\cdots,B_r,\{a_1\},\cdots,\{a_l\}\}$ in $T_n$.
Therefore we have that $\color{blue}{B_n=C_n+C_{n+1}}$ for $n\ge1$; so 
$\;C_n=B_{n-1}-C_{n-1}=B_{n-1}-\left(B_{n-2}-C_{n-2}\right)= \;\cdots\;=B_{n-1}-B_{n-2}+\cdots+(-1)^nB_1$.
A: The exponential generating function for $B_n$ (the Bell numbers) is given by
$$
B(x)=\sum_{n=0}^{\infty}B_n\frac{x^n}{n!}=\exp(e^x-1)\tag{1}
$$
since the Bell-numbers satisfy the recurrence
$$
B_{n+1}=\sum_{k=0}^{n}\binom{n}{k} B_k;\quad B_0=B_1=1.
$$
The exponential generating function for $C_n$ is given by
$$
C(x)=\sum_{n=0}^{\infty}C_n\frac{x^n}{n!}=\exp(e^x-1-x).\tag{2}
$$
Note that
$$
C(x)+C'(x)=B(x)
$$
which implies that
$$
B_n=C_n+C_{n+1}. \quad(n\geq 1)
$$
At this point look at the last line of the answer of user84413.
