Bernoulli numbers explicit form I found this nice explicit formula for the Bernoulli numbers:
$$B_n = \sum_{k \mathop = 0}^n \sum_{i \mathop = 0}^k (-1)^i \binom k i \frac {i^n} {k + 1}$$
I can't find a proof though. I want to prove it from the generating function definition:
$$ \frac x {e^x - 1} = \sum_{n \mathop = 0}^\infty B_n \frac {x^n} {n!}$$
Any proof sketches or links will be appreciated.
 A: So as not  to get mixed up with complex variables  use $j$ rather than
$i$ to get
$$\sum_{k=0}^n \sum_{j=0}^k (-1)^j {k\choose j} \frac{j^n}{k+1}.$$
This is
$$\sum_{k=0}^n \frac{1}{k+1} \sum_{j=0}^k (-1)^j {k\choose j} j^n
= \sum_{k=0}^n \frac{1}{k+1} (-1)^k \times k! \times {n\brace k}.$$
Recall the classic generating function  of the Stirling numbers of the
second kind which yields
$${n\brace k} = n! [z^n][u^k] \exp(u(\exp(z)-1)).$$
Substituting this into the sum gives
$$n![z^n] 
\sum_{k=0}^n \frac{1}{k+1} (-1)^k \times k! \times 
[u^k] \exp(u(\exp(z)-1))
\\ = n![z^n] 
\sum_{k=0}^n \frac{1}{k+1} (-1)^k \times k! \times 
\frac{(\exp(z)-1)^k}{k!} 
\\= n![z^n] 
\sum_{k=0}^n \frac{1}{k+1} (-1)^k \times 
(\exp(z)-1)^k$$
Now observe that $\exp(z)-1$ starts at $z$ and hence we can extend the summation to infinity without affecting $[z^n]$ to get
$$n![z^n] 
\sum_{k=0}^\infty \frac{1}{k+1} (-1)^k \times 
(\exp(z)-1)^k
\\=n! [z^n]
\frac{1}{\exp(z)-1}
\sum_{k=0}^\infty \frac{1}{k+1} (-1)^k \times 
(\exp(z)-1)^{k+1}
\\= n! [z^n] \frac{1}{\exp(z)-1}
\log(1+\exp(z)-1)
= n! [z^n] \frac{z}{\exp(z)-1}.$$
Done.
Nice how Bernoulli numbers show up in both analytic number theory and combinatorics.
A: There have been many references collected for proofs of this formula of the Bernoulli numbers $B_n$ at https://math.stackexchange.com/a/4254493/945479.
