Recurrence Relation of Bernoulli Numbers. I am trying to understand the Recurrence Relation of Bernoulli Numbers. 
It is defined that, 
$$
s_{p}{(n)} = \sum_{k=0}^{p}\frac{B_k}{k!}\frac{p!}{(p+1-k)!}n^{p+1-k} \tag 1
$$
Then, for $n$ = $1$, equation($1$) comes as, 
$$
0 = \sum_{k=0}^{p}\frac{B_k}{k!}\frac{p!}{(p+1-k)!} \tag 2
$$
Similarly, for $n$=$p$, equation($1$) becomes, 
$$
B_p = -\frac{1}{p+1}\sum_{k=0}^{p-1}\binom{p+1}{k}B_k \tag 3
$$
Later, the above relation (at $(3)$) is re-written symbolically as, 
$$
B^{p+1}-B^{p}=0 \tag 4
$$
My doubt is, how does the equation come down from $(1)$ to $(3)$ (for $n$=$p$) and then from $(3)$ to $(4)$ ?
I am following the tutorial at this link for my study; and the above equations and relations are from section 4, sub-section 4.1 of the tutorial.
I am a beginner, any kind of help or suggestion is very much appreciated.
Regards. :)
 A: The text you refer to defines
$$s_p (n) = \sum_{1 \le k \le n-1} k^p.$$
(You should have mentioned that!) In particular, note that according to this definition, $s_p (1) = 0$.
One can see that this function is a polynomial in $n$, and its coefficients may be declared to be the Bernoulli numbers, up to some factors (Bernoulli himself discovered $B_k$ while studying $s_p (n)$). Namely, your formula (1) may be written as
$$\tag{1} s_p (n) = \frac{1}{p+1} \sum_{0\le k \le p} {p+1 \choose k}\,B_k\,n^{p+1-k}.$$
If $n = 1$, then the above equation becomes
$$\tag{2} 0 = \frac{1}{p+1}\,\sum_{0\le k \le p} {p+1 \choose k}\,B_k,$$
that is,
$$\tag{4} \sum_{0\le k \le p} {p+1 \choose k}\,B_k = 0.$$
This recurrence is for $p \ge 1$, and the initial value is $B_0 = 1$.
In particular, the coefficient of $B_p$ is ${p+1 \choose p} = p+1$, and the expression for $B_p$ is
$$\tag{3} B_p = -\frac{1}{p+1}\,\sum_{0\le k \le p-1} {p+1 \choose k}\,B_k.$$
The formula "$(B+1)^{p+1} - B^{p+1} = 0$" apparently means that you should expand the term $(B+1)^{p+1}$ via the binomial theorem:
$$(B+1)^{p+1} - B^{p+1} = \sum_{0 \le k \le p} {p+1 \choose k} B^k$$
and then replace $B^k$ with $B_k$. This is just a fancy way to express the recurrence relation (4).

Usually one defines $s_p (n)$ to be $\sum_{1 \le k \le n} k^p$. Then $s_p (1) = 1$, and you will end up with the recurrence
$$\sum_{0\le k \le p} {p+1 \choose k}\,B_k = p+1.$$
This would give $B_0 = 1, B_1 = \frac{1}{2}$, while your definition gives $B_0 = 1, B_1 = -\frac{1}{2}$.
