Identity of Bernoulli polynomials I am trying to prove the following identity: $$B_n(1-x)=(-1)^nB_n(x)$$  I know that $$B_n(1-x)=\sum_{k=0}^n\binom n k B_n\cdot (1-x)^{n-k}$$ (I do not know how to write here exactly what I want. So I mean that $(1-x)$ in degree $(n-k)$).  Hence $$B_n(1-x)=\sum_{k=0}^n\binom n k B_n \sum_{i=0}^{n-k}\binom{n-k}{i} (-1)^ix^i$$  Hope, you understand what I mean. Thanks for attention!
Proving an identity for Bernoulli polynomials Consider the Bernoulli polynomials $B_n(x)$ given by the expansion  $$\frac{te^{xt}}{e^t-1} = \sum\limits_{n=0}^{\infty}B_n(x)\frac{t^n}{n!}.$$  I want to prove the identity $$B_n(1-x)=(-1)^nB_n(x).$$  This formula is given in a lot of books, but unfortunately without any proof. Therefore I'm wondering how to prove it. Do you have any idea?  I tried for instance:  $\sum\limits_{n=0}^{\infty}B_n(1-x)\frac{t^n}{n!}=\frac{te^{-xt}}{e^t-1}e^t=\left(\sum\limits_{n=0}^{\infty}B_n(-x)\frac{t^n}{n!}\right)\sum\limits_{n=0}^{\infty}\frac{t^n}{n!}=\sum\limits_{n=0}^{\infty}\left(\sum\limits_{l=0}^n\binom{n}{l}B_{l}(-x)  \right)\frac{t^n}{n!}$.  Hence  $B_n(1-x)=\sum\limits_{l=0}^n\binom{n}{l}B_{l}(-x)$.  But I do not know how to proceed.  On the other hand I know that $B_n(x+1)-B_n(x)=n x^{n-1}$. Hence  $B_{n}(1-x)=B_{n}(-x)-n(-x)^{n-1}$. But again, I do not know how I can go further.   Hopefully someone can help me.  Best regards
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