<s>[INST] 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.
 [/INST] 
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.

</s>