Proof Bell-Number $B(n+1)=\sum\limits_{i=0}^n\binom{n}{i}B(i)$ Let B(0) := 1 und B(n) for n$\geq$1 the counts of all sets partitions of [n]. The numbers B(n) are the Bell-numbers.
For $n \geq 0$ prove that: 
\begin{equation}
B(n+1)=\sum\limits_{i=0}^n\binom{n}{i}B(i)
\end{equation}


*

*I think to construct a subset of i-elements $A_i$ of [n] 

*split $A_i$ in k not empty and disjunct blocks $B_1\dots B_k$ where $A_i= \bigcup \limits_{i=1}^k B_i$

*We have another k+1 block $B_{k+1}=[n]$ \ $A_i \cup \{n+1\}$
So we have a set of partion of [n+1] in k+1 portions


for 1: there are $\binom{n}{i}$ ways
for 2: there are B(n) ways
if we sum on n then we become $B(n+1)=\sum\limits_{i=0}^n\binom{n}{i}B(i)$
What is wrong in this proof ?
 A: For reference here is a proof using generating functions. 

Observe that when we  multiply two exponential generating functions of
the sequences $\{a_n\}$ and $\{b_n\}$ we get that
$$ A(z) B(z) = \sum_{n\ge 0} a_n \frac{z^n}{n!}
\sum_{n\ge 0} b_n \frac{z^n}{n!}
= \sum_{n\ge 0}
\sum_{k=0}^n \frac{1}{k!}\frac{1}{(n-k)!} a_k b_{n-k} z^n\\
= \sum_{n\ge 0}
\sum_{k=0}^n \frac{n!}{k!(n-k)!} a_k b_{n-k} \frac{z^n}{n!}
= \sum_{n\ge 0}
\left(\sum_{k=0}^n {n\choose k} a_k b_{n-k}\right)\frac{z^n}{n!}$$
i.e. the  product of  the two generating  functions is  the generating
function of $$\sum_{k=0}^n {n\choose k} a_k b_{n-k}.$$

The species of set partitions has specification $$\mathfrak{P}(\mathcal{U}\mathfrak{P}_{\ge 1}(\mathcal{Z}))$$ which gives the generating function
$$G(z, u) = \exp(u(\exp(z)-1)).$$
Therefore the EGF of Bell numbers is $$B(z)=\exp(\exp(z)-1)
= \sum_{n\ge 0} B_n\frac{z^n}{n!}.$$
Differentiating we obtain
$$\sum_{n\ge 1} B_n\frac{z^{n-1}}{(n-1)!}
= \sum_{n\ge 0} B_{n+1}\frac{z^n}{n!}
= \exp(\exp(z)-1)\exp(z).$$
By convolution of generating functions this immediately implies
$$B_{n+1}=\sum_{q=0}^n {n\choose q} B_q.$$
