I have pored over the internet and can't seem to find any proof of a particular property of Bernoulli Polynomials. The property is:
$P'_{n}(x) = nP_{n-1}(x)$
I have not found anything conclusive in any of my textbooks that lay out a clear proof or anything on the internet. Any help or resources as to what the proof might be would be greatly appreciated!
From the definition of bernoulli polynomials, we have
$$B_n(x)=\sum_{k=0}^n\binom{n}{k}B_{n-k}x^k$$
where $B_k$ are the traditional Bernoulli numbers. Differentiate both sides to get
\begin{eqnarray*}B'_n(x)&=&\frac{d}{dx}\sum_{k=0}^n\binom{n}{k}B_{n-k}x^k\\&=&\sum_{k=1}^n\binom{n}{k}kB_{n-k}x^{k-1}\\&=&\sum_{k=0}^{n-1}\binom{n}{k+1}(k+1)B_{n-k+1}x^k\\&=&n\sum_{k=0}^{n-1}\binom{n-1}{k}B_{n-1-k}x^k\\&=&nB_{n-1}(x)\end{eqnarray*}
Where we used the combinatorial identity
$$(k+1)\binom{n}{k+1}=n\binom{n-1}{k}$$