prove for bell number using induction on n hi guys I have to prove this equality $$B_n=e^{-1}\sum_{k=0}^{\infty}\frac{k^n}{k!},$$ that is called bell equality only using induction on $n$ . How can i do this? I have tried by  substituting the recursive formula  $\sum\limits_{k=0}^{n} \binom{n}{k} B_{k}$ in the first one. But I am completely lost with indexes.
I have already tried to search some informations on the internet but nothing using induction
 A: Using the Bell´s formula we have that $$B(n)=\sum\limits_{k=0}^nS(n,k).$$
This is the total number of ways to put $n$ balls in an arbitrary number of
boxes (no empty boxes remaining). To count them we look at the number of balls (at this parameter
we will call it $a$, and it can be from $0$ to $(n - 1)$) that accompany any ball (for example ball
number $1$) on your box. To do this:

*

*First we choose the $a$ companions of ball $1$ from among the $(n - 1)$ possible (which can be done in $\binom{n-1}{a}$ different ways).

*Then, for each way of selecting the companions, we count how many ways they can be entered
the remaining $(n − 1 − a)$ balls in an arbitrary number of boxes (we have called this $B(n−1−a)$)

$$\boxed{B(n)}=\sum\limits_{a=0}^{n-1}{\dbinom{n-1}{a}}\sum\limits_{k=1}^{n-1-a}{S(n-1-a)}=\boxed{\sum\limits_{a=0}^{n-1}{\dbinom{n-1}{a}B(n-1-a)}}$$
Now, if we define the $DOB$ function this way: $$DOB(n)=\sum\limits_{k=1}^{\infty}\dfrac{k^n}{n!},$$ we can see that
\begin{align*}
DOB(0)&=1+\dfrac{1}{1!}+\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+\cdots\\
DOB(1)&=\phantom{OO}\dfrac{1}{1!}+\dfrac{2}{2!}+\dfrac{3}{3!}+\dfrac{4}{4!}+\cdots\\
DOB(2)&=\phantom{OO}\dfrac{1^2}{1!}+\dfrac{2^2}{2!}+\dfrac{3^2}{3!}+\dfrac{4^2}{4!}+\cdots\\
DOB(3)&=\phantom{OO}\dfrac{1^3}{1!}+\dfrac{2^3}{2!}+\dfrac{3^3}{3!}+\dfrac{4^3}{4!}+\cdots
\end{align*}
By combining these numbers we get that
\begin{gather*}
\color{red}{1}DOB(0)+\color{red}{2}DOB(1)+\color{red}{1}DOB(2)=1+\dfrac{4}{1!}+\dfrac{9}{2!}+\dfrac{27}{3!}\cdots=\dfrac{1^3}{1!}+\dfrac{2^3}{2!}+\dfrac{3^3}{3!}+\dfrac{4^3}{4!}+\cdots=DOB(3)\\
\color{red}{1}DOB(0)+\color{red}{3}DOB(1)+\color{red}{3}DOB(2)+\color{red}{1}DOB(3)=1+\dfrac{8}{1!}+\dfrac{27}{2!}+\dfrac{64}{3!}\cdots=\dfrac{1^4}{1!}+\dfrac{2^4}{2!}+\dfrac{3^4}{3!}+\dfrac{4^4}{4!}+\cdots=DOB(4)
\end{gather*}
What Dobinsky wanted to prove is that both sequences fulfill the same recurrence, and that have the following property:
$$DOB(n)=\sum\limits_{j=0}^{n-1}{\dbinom{n-1}{j}DOB}(j)$$
To prove this just use the binomial Theorem:
\begin{align*}
 \boxed{\sum\limits_{j=0}^{n-1}{\dbinom{n-1}{j}DOB(j)}}=&\ e+\sum\limits_{j=1}^{n-1}{\dbinom{n-1}{j}DOB(j)}=e+\sum\limits_{j=1}^{n-1}{\dbinom{n-1}{j}\sum\limits_{k=1}^{\infty}{\dfrac{k^n}{n!}}}\\
 =&\ e+\sum\limits_{k=1}^{\infty}{\dfrac{1}{k!}}\underbrace{\sum\limits_{j=1}^{n-1}{\dbinom{n-1}{j}k^j}}_{=(k+1)^{n-1}-1}=\ e-\underbrace{\sum\limits_{k=1}^{\infty}{\dfrac{1}{k!}}}_{=e-1}+\sum\limits_{k=1}^{\infty}{\dfrac{(k+1)^{n-1}}{k}}\\
 =&1+\sum\limits_{k=1}^{\infty}{\dfrac{(k+1)^{n-1}}{k!}=1+\sum\limits_{k=1}^{\infty}{\dfrac{(k+1)^n}{(k+1)!}}=\sum\limits_{k=1}^{\infty}{\dfrac{j^n}{j!}}}=\boxed{DOB(j)}
\end{align*}
This tells us that:
\begin{equation}\label{Formula_Dobinsky}
eB(n)=DOB(n) \rightarrow \boxed{B(n)=\dfrac{1}{e}\sum\limits_{k=1}^{\infty}{\dfrac{k^n}{k!}}}\quad \text{for each }n=1,2,3\ldots
\end{equation}
This formula is way more friendly and computerwise more stable. To see that just check that
\begin{equation*}
B(10)=115975;\qquad\dfrac{1}{e}\sum\limits_{k=1}^{15}{\dfrac{k^{10}}{k!}}=115974.978
\end{equation*}
