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
Hint. You may just write $$ \frac{te^{(1-x)t}}{e^t-1}=\frac{t\:e^{t}e^{x(-t)}}{e^t(1-e^{-t})}=\frac{t\:e^{x(-t)}}{1-e^{-t}}=\frac{(-t)e^{x(-t)}}{e^{(-t)}-1} $$ and use the definition.