question on summation? Please, I need to know the proof that 
$$\left(\sum_{k=0}^{\infty }\frac{n^{k+1}}{k+1}\frac{x^k}{k!}\right)\left(\sum_{\ell=0}^{\infty }B_\ell\frac{x^\ell}{\ell!}\right)=\sum_{k=0}^{\infty }\left(\sum_{i=0}^{k}\frac{1}{k+1-i}\binom{k}{i}B_in^{k+1-i}\right)\frac{x^k}{k!}$$
where $B_\ell$, $B_i$ are Bernoulli  numbers.
Maybe we should replace $k$ with $j$?
Anyway, I need to prove how to move from the left to right. 
Thanks for all help.

The right-hand side is just the Cauchy product of the two series on the left-hand side. Generally, the Cauchy product of $\sum_{k=0}^\infty a_k$ and $\sum_{l}^\infty b_l$ is $$\sum_{j=0}^\infty\sum_{i=0}^j b_i a_{j-i}.$$
 Now you should have no problem finishing the problem. (Change $j$ to $k$ at the end to match your right hand side.)
