how the following formula or any other explicit formula for computing Bernoulli numbers can be derived? how the following formula 
$$B_n=1-\sum_{k=0}^{n-1}{n\choose k}\frac{B_k}{n-k+1}$$
can be derived?
I know how to use the formula but still have not seen any proof for the given formula.
 A: It is immediate from the definition $$\frac{x}{e^x-1} = \sum_{k=0}^\infty \frac{B_k}{k!} x^k$$ which means $$1=\frac{e^x-1}{x} \sum_{k=0}^\infty \frac{B_k}{k!} x^k = ( \sum_{m=0}^\infty \frac{1}{(m+1)!} x^m)( \sum_{k=0}^\infty \frac{B_k}{k!} x^k) =  \sum_{n=0}^\infty (\sum_{k=0}^n \frac{B_k}{k!}\frac1{ (n-k+1)!}) x^n$$
ie. $$B_0=1, \qquad  for \ n > 1\ \qquad \sum_{k=0}^n \frac{B_k}{k! (n-k+1)!} = 0$$
A: The following formula
 has been proved in The American Mathematical Monthly, but here the proof is deeper and doesn't use any induction:
consider the function $$f\left(x\right)=\frac{1}{e^{x}+1}=\frac{e^{x}+1-2}{e^{2x}-1}=\frac{1}{e^{x}-1}-\frac{2}{e^{2x}-1}=\frac{1}{x}\cdot\frac{x}{e^{x}+1}-\frac{1}{x}\cdot\frac{2x}{e^{2x}-1}$$
Using the famous definition of Bernoulli numbers we have:
$$f\left(x\right)=\frac{1}{x}\sum_{n=0}^{∞}B_{n}\ \frac{x^{n}}{n!}-\frac{1}{x}\sum_{n=0}^{∞}B_{n}\ \frac{2^{n}x^{n}}{n!}=\sum_{n=0}^{∞}B_{n}\ \frac{x^{n-1}}{\left(n\right)!}\left(1-2^{n}\right)$$
let $n-1↦n$, implies:$$f\left(x\right)=\sum_{n=0}^{∞}B_{n+1}\ \frac{x^{n}}{\left(n+1\right)!}\left(1-2^{n+1}\right)$$
Computing $$f^{\left(n\right)}\left(0\right)$$ gives the following relation:
(I)
$$f^{\left(n\right)}\left(0\right)=\frac{B_{n+1}\left(1-2^{n+1}\right)}{n+1}$$ 
On the other hand from the answer of user90369 we can compute the $n^{th}$ derivative of $$ \frac{1}{e^{x}+1}$$ around $x=0$ which is as follows:
(II) $$\frac{d^{n}}{dx^{n}}\frac{1}{e^{x}+1}=\sum_{k=0}^{n}\left(-1\right)^{n}\sum_{j=0}^{k}\left(-1\right)^{j}{{k}\choose{j}}\left(j+1\right)^{n}\cdot\frac{1}{2^{k+1}}$$ 
comparing the $n^{th}$  derivative of the function from I and II we arrive at the explicit formula for the $n+1^{th}$ Bernoulli number:
$$B_{n+1}=\frac{\left(-1\right)^{\left(n+1\right)}\left(n+1\right)}{2^{\left(n+1\right)}-1}\sum_{k=1}^{n+1}\frac{1}{2^{\left(k\right)}}\sum_{j=0}^{k-1}\left(-1\right)^{j}\left(j+1\right)^{n}{{k-1}\choose{j}}$$ which is valid for $n\ge0$.
A: 
The recurrence relation 
  \begin{align*}
B_n&=1-\sum_{k=0}^{n-1}{n\choose k}\frac{B_k}{n-k+1}\qquad n\geq 0\\
\end{align*}
specifies a variation of the Bernoulli numbers with $B_1=+\frac{1}{2}$ instead of the more common value $B_1=-\frac{1}{2}$. 
We therefore start with the generating function
  \begin{align*}
\sum_{k=0}^\infty \frac{B_k}{k!}x^k=\frac{x}{e^x-1}+x=\frac{xe^x}{e^x-1}\tag{1}
\end{align*}

Note: The recurrence relation also gives $B_0=1$ since we use the convention that an empty sum is equal to zero. This is the case for $n=0$ when the upper limit of the sum is less than the lower limit.
Multiplication of (1) by $e^x-1$ gives
\begin{align*}
xe^x&=\left(\sum_{k=0}^\infty\frac{B_k}{k!}x^k\right)\left(e^x-1\right)\tag{2}\\
&=\left(\sum_{k=0}^\infty\frac{B_k}{k!}x^k\right)\left(\sum_{l=1}^\infty\frac{x^l}{l!}\right)\tag{3}\\
&=\sum_{n=1}^\infty \left(\sum_{{k+l=n}\atop{k\geq 0,l\geq 1}}\frac{B_k}{k!}\frac{1}{l!}\right)x^n\tag{4}\\
&=\sum_{n=1}^\infty\left(\sum_{k=0}^{n-1}\frac{B_k}{k!}\frac{1}{(n-k)!}\right)x^n\tag{5}\\
&=\sum_{n=0}^\infty\left(\sum_{k=0}^n\frac{B_k}{k!}\frac{1}{(n-k+1)!}\right)x^{n+1}\tag{6}\\
&=\sum_{n=0}^\infty\left(\sum_{k=0}^n\binom{n}{k}\frac{B_k}{n-k+1}\right)\frac{x^{n+1}}{n!}\tag{7}\\
\end{align*}
Comment:


*

*In (3) we expand the exponential function $e^x=\sum_{n=0}^\infty \frac{x^n}{n!}$.

*In (4) we do the Cauchy multiplication of series.

*In (5) we eliminate $l$ by using $l=n-k$.

*In (6) we shift the index $n$ to start with $n=0$.

*In (7) we expand numerator and denominator by $n!$ and introduce the binomial coefficient $\binom{n}{k}$.

From (7) and the expansion of the left-hand side of (2) we obtain after division by $x$
\begin{align*}
\sum_{n=0}^\infty \frac{x^n}{n!}=\sum_{n=0}^\infty\left(\sum_{k=0}^n\binom{n}{k}\frac{B_k}{n-k+1}\right)\frac{x^{n}}{n!}\tag{8}
\end{align*}
Comparison of the coefficient of $x^n$ gives after multiplication with $n!$ for $n\geq 0$:
\begin{align*}
\color{blue}{1}&=\sum_{k=0}^n\binom{n}{k}\frac{B_k}{n-k+1}\\
&\,\,\color{blue}{=B_n+\sum_{k=0}^{n-1}\binom{n}{k}\frac{B_k}{n-k+1}}
\end{align*}
and the claim follows.

