Proving of the multiplication theorem for Bernoulli polynomial How the expression below can be proven:
$$B_n(mx) = m^{n−1} \sum\limits_{k=0}^{m-1}B_n\left(x+\frac{k}{m}\right)$$
Where $B_n(x)$ is Bernoulli polynomial
I know it is already proved by Joseph Ludwig Raabe, but I don`t know how exactly. 

By induction over $n$.
If $n=1$ we have from definition of $B_1(x)=x-1/2$,
\begin{eqnarray*}
  B_1(mx) = mx - \frac{1}{2},
\end{eqnarray*}
On the other hand, the right hand side with $n=1$ is
\begin{eqnarray*}
  \sum_{k=0}^{m-1} B_1 \left (  x +\frac{k}{m} \right ) 
  &=& \sum_{k=0}^{m-1} \left (  x+ \frac{k}{m} - \frac{1}{2}  \right ) \\
  &=& mx + \frac{1}{m} \frac{m(m-1)}{2} - \frac{m}{2} \\
  &=& mx - \frac{1}{2}
\end{eqnarray*}
Assume that for $n-1$ the equality holds.
Define $f_n(x) = B_n(mx) - m^{n-1} \sum_{k=0}^{m-1} B_n \left ( x + \frac{k}{m} \right )$.
Since $dB_n(x)/dx= n B_{n-1}(x)$, 
The derivative of $f_n$ is given by
\begin{eqnarray*}
  f'_n(x) &=& m n B_{n-1}(mx)  - m^{n-1} \sum_{k=0}^{m-1} \frac{n}{m} B_{n-1} \left (
  x + \frac{k}{m} \right ) \\
  &=& m n f_{n-1}(x) \\
  &=& 0   \quad \text { by the induction hypothesis on $n-1$ }.
\end{eqnarray*}
Hence $f_n(x)=c_n$ for $c_n$ constant. To find the constant 
note that since $\int_0^1 B_n(x) dx = 0$, then 
\begin{eqnarray*} \int_0^{1/m} f_n(x) &=& 
  \int_0^{1/m}  \left [ B_n(mx) - m^{n-1} \sum_{k=0}^{m-1} B_n \left ( x + \frac{k}{m} \right )
  \right ] dx \\
  &=& \frac{1}{m} \int_0^1   B_n(y) dy  - 
  m^{n-1} \sum_{k=0}^{m-1} \int_0^{1/m} B_n \left ( x + \frac{k}{m} \right ) dx \\
  &=& 0 + m^{n-1} \sum_{k=0}^{m-1} \int_{k/m}^{(k+1)/m} B_n \left ( y \right ) dy \\
  &=& 0 + m^{n-1} \sum_{k=0}^{m-1} \int_0^1 B_n \left ( y \right ) dy \\
  &=& 0. 
\end{eqnarray*}
So, since the integral of a constant in the interval $[0,1/m]$, $m > 0$, is 0, 
that constant has to be zero. Hence $f_n(x)=0$ as desired.
