Bernoulli numbers. Today I read a short article about Bernoulli numbers. This article give the way to calculate Bernoulli numbers.
$\displaystyle \frac{x}{e^{x}-1} = f(x)$
$\displaystyle f(x)(e^{x}-1) = x = \sum_{n = 0}^{\infty}B_{n} \frac{x^{n}}{n!} \sum_{k = 1}^{\infty} \frac{x^{k}}{k!} = \sum_{n=1}^{\infty}x^{n} \sum_{k+m = n}^{\infty}\frac{B_{m}}{k!m!} = \sum_{n=1}^{\infty} \frac{x^{n}}{n!} \sum_{m=0}^{n-1} \binom{n}{m} B_{m}$
Now we could calculate Bernoulli numbers for different $n$. But I don't understand the step with $k+m = n$. Please help me with it!

In slow motion
$$
\begin{array}{l}
 \sum\limits_{0\, \le \,n} {B_{\,n} \frac{{x^{\,n} }}{{n!}}} \sum\limits_{1\, \le \,k} {\frac{{x^{\,k} }}{{k!}}}  = \sum\limits_{\begin{array}{*{20}c}
   {0\, \le \,n}  \\
   {1\, \le \,k}  \\
\end{array}} {B_{\,n} \frac{{x^{\,n + k} }}{{n!k!}}}  = \sum\limits_{\begin{array}{*{20}c}
   {0\, \le \,n}  \\
   {n + 1\, \le \,n + k}  \\
\end{array}} {B_{\,n} \frac{{x^{\,n + k} }}{{n!k!}}}  =  \\ 
  = \sum\limits_{\begin{array}{*{20}c}
   {0\, \le \,n}  \\
   {n + 1\, \le \,q}  \\
\end{array}} {B_{\,n} \frac{{x^{\,q} }}{{n!\left( {q - n} \right)!}}}  = \sum\limits_{\begin{array}{*{20}c}
   {1\, \le \,q}  \\
   {0\, \le \,n\, \le \,q - 1}  \\
\end{array}} {B_{\,n} \frac{{x^{\,q} }}{{n!\left( {q - n} \right)!}}}  =  \\ 
  = \sum\limits_{\begin{array}{*{20}c}
   {1\, \le \,q}  \\
   {0\, \le \,n\, \le \,q - 1}  \\
\end{array}} {B_{\,n} \frac{{q!}}{{n!\left( {q - n} \right)!}}\frac{{x^{\,q} }}{{q!}}}  = \sum\limits_{1\, \le \,q} {\left( {\sum\limits_{0\, \le \,n\, \le \,q - 1} {B_{\,n} \left( \begin{array}{l}
 q \\ 
 n \\ 
 \end{array} \right)} } \right)\frac{{x^{\,q} }}{{q!}}}  =  \\ 
  = \sum\limits_{1\, \le \,n} {\left( {\sum\limits_{0\, \le \,m\, \le \,n - 1} {B_{\,m} \left( \begin{array}{l}
 n \\ 
 m \\ 
 \end{array} \right)} } \right)\frac{{x^{\,n} }}{{n!}}}  \\ 
 \end{array}
$$
