Bernoulli numbers explicit form

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.