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!
The Bernoulli polynomials are defined by the exponential generating function $$ g(x,t) = \frac{t e^{tx}}{e^t-1} = \sum_{n=0}^\infty \dfrac{B_n(x) t^n}{n!} $$ Your equation $B_n(1-x) = (-1)^n B_n(x)$ comes from $$ g(1-x, t) = g(x,-t)$$ which is easy to verify.