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.

$$\left(\sum_{k=0}^{\infty} \dfrac{n^{k+1}}{k+1} \dfrac{x^k}{k!} \right) \left(\sum_{l=0}^{\infty} B_l \dfrac{x^l}{l!}\right) = \sum_{k,l} \dfrac{n^{k+1}}{k+1} \dfrac{B_l}{k! l!} x^{k+l}$$
$$\sum_{k,l} \dfrac{n^{k+1}}{k+1} \dfrac{B_l}{k! l!} x^{k+l} = \sum_{m=0}^{\infty} \sum_{l=0}^{m} \dfrac{n^{m-l+1}}{m-l+1} \dfrac{B_l}{(m-l)! l!} x^{m}$$
This gives us
$$\sum_{m=0}^{\infty} \sum_{l=0}^{m} \dfrac{B_l}{(m-l+1)! l!}n^{m-l+1} x^{m} = \sum_{m=0}^{\infty} \left(\sum_{l=0}^m \dfrac1{m-l+1} \dbinom{m}{l} B_l n^{m-l+1}\right)\dfrac{x^m}{m!}$$
