Does anyone recognise this recursion sastisfied by the Bell numbers? I derived a recursion 
$$B_n=\frac{1}{n}B_0+\frac{1}{n}\sum_{k=1}^{n-1}\left[\binom{n}{k}-(-1)^{n-k}\binom{n}{k-1}\right]B_k\tag{$*$}$$
which I know should be satisfied by the moments of the unit Poisson distribution ($B_n$ denoting the $n$th moment), which happen to be the Bell numbers. Trying to check that I hadn't mucked up the derivation I looked online for recursions satisfied by the Bell numbers hoping to find the above, however I only found the standard one
$$B_n = \sum_{k=0}^{n-1}\binom{n-1}{k}B_k.$$ 
I'm pretty sure now that the derivation of $(*)$ is fine (I've also computed the first few numbers obtained from $(*)$, with $B_0=1$, and they are the first few Bell numbers). So does anyone recognise the recursion $(*)$? Does it have a name, or, do you know of a reference that contains it? Thanks.
Edit: Happy to give the bounty to anyone that posts a proof that $(*)$ is (or isn't) satisfied by the Bell numbers or a reference containing such a proof.
 A: Since I am very lazy, I just computed $$R_n=\frac{1}{n}B_0+\frac{1}{n}\sum_{k=1}^{n-1}\left[\binom{n}{k}-(-1)^{n-k}\binom{n}{k-1}\right]B_k-\sum_{k=0}^{n-1}\binom{n-1}{k}B_k$$ and effectively found that $R_n=0$ for any positive value of $n$
A: It is known that the Bell number $B_n$ satisfy a recurrence relation involving binomial coefficients:
$$B_{n+1} = \sum_{k=0}^n \binom{n}{k} B_k \quad\text{ for } n \in \mathbb{N}\tag{*1}.$$
Given any sequence $( a_n )_{n \in \mathbb{N}}$, we can generate
another sequence $(b_n)_{n \in \mathbb{N}}$ by following transform:
$$b_n = \sum_{k=0}^n (-1)^{n-k}\binom{n}{k} a_k \tag{*2a}$$
This is known as the binomial transform
and it has an inverse transform of the form:
$$a_n = \sum_{k=0}^n \binom{n}{k} b_k\tag{*2b}$$
Compare $(*1)$ with $(*2b)$, the recurrence relation $(*1)$ is really the inverse transform
for the relation:
$$\begin{align}
B_n &= \sum_{k=0}^n (-1)^{n-k}\binom{n}{k} B_{k+1}
= B_{n+1} - n B_n - \sum_{k=1}^{n-1} (-1)^{n-k} \binom{n}{k-1} B_k\\
\iff
n B_n &= (B_{n+1} - B_n) - \sum_{k=1}^{n-1} (-1)^{n-k} \binom{n}{k-1} B_k
\end{align}
$$
Substitute $(*1)$ into last expression, we get
$$\begin{align}
n B_n 
&= B_0 + \sum_{k=1}^{n-1} \binom{n}{k} B_k
- \sum_{k=1}^{n-1} (-1)^{n-k} \binom{n}{k-1} B_k\\
&= B_0 + \sum_{k=0}^{n-1}\left[ \binom{n}{k} - (-1)^{n-k}\binom{n}{k-1}\right] B_k
\end{align}
$$
So the recurrence relation in the question is nothing really new but the recurrence relation $(*1)$ hidden behind a binomial transform.
