I got a little question regarding the Bernoulli polynomials/numbers. Basically, I want to show that $$B_n(0) = -\frac{B_n(2\pi i)^n}{n!}$$ Where $B_n(x)$ the $n$-th Bernoulli polynomial and $B_n$ the $n$-th Bernoulli number). I have gotten quite far, I think, but now I seem to be stuck:
I got that $\sum_{k=1}^n B_{n-k}(0)\frac{(2\pi i)^k}{k!}=0=\sum_{k=1}^n\frac{B_{n-k}}{(n-k)!k!}$. Now, I read up on it a bit, and usually what I find is something along the lines of "we compare the coefficients and find the above identity". However, I don't see how we can arrive at that conclusion (I see that we can reindex the sums, such that it looks about right though). My main problem is: how can we know that if we look at the k-th component in the sum, they have to match up. Silly example, but $1+2+3=3+2+1$, there, the sum is the same, but the elements don't match up (if you don't reorganize them).
Can somebody help me out, I'd appreciate it!