Bernoulli Numbers generating function and Riemann Zeta function

Question

I've been studying Bernoulli numbers and I came across this summation: $$ \sum_{n=1}^{\infty}\frac{B_n x^n}{n!} = \sum_{n=1}^{\infty}\frac{-n \zeta(1-n) x^n}{n!} = -\sum_{n=1}^{\infty}\frac{\zeta(1-n) x^n}{(n-1)!} = -\sum_{n=1}^{\infty}\frac{x^n}{(n-1)!} \sum_{k=1}^\infty k^{n-1} =$$ $$ -\sum_{n=1}^{\infty}\sum_{k=1}^\infty\frac{x x^{n-1} k^{n-1}}{(n-1)!} = -x\sum_{n=1}^{\infty}\sum_{k=1}^\infty\frac{(x k)^{n-1}}{(n-1)!} = -x\sum_{n=1}^\infty \exp(x k) = \frac{x e^x}{e^x-1}$$ which doesn't work. (It should be $\frac{x}{e^x-1}$). Why?

Answer 1

I think I solved it. The Zeta function and it's equation for Bernoulli numbers is for $B_n(1)$, not $B_n(0)$. And the generating function for $B_t(x)$ is $\frac{t e^{x t}}{e^t-1}$. So it works.