Proof that $\zeta (-n)=(-1)^n\frac{B_{n+1}}{n+1}$ for $n\ge 0$

Question

I'd like to prove that $$\zeta (-n)=(-1)^n\dfrac{B_{n+1}}{n+1}\quad n\ge 0.$$ The only approach I found uses $$\zeta (-n)=\dfrac{in!}{2\pi}\displaystyle\oint_C \dfrac{(-t)^{-n-1}}{e^t-1} \, dt.$$ Can the identity for $\zeta (-n)$ be proved by any other means (not by contour integration)? At least links to some sources would be helpful.

Answer 1

One method is to use the functional equation. Dr. Jerry Shurman sketches this in Zeta at Negative Odd Integers, a la Euler. First he shows, for even $k \geq 2$,
$$ \zeta(1-k) = \frac{2 \Gamma(k)}{(2 \pi \mathrm{i})^k} \zeta(k) $$ This ties values of $\zeta$ at negative odd integers to reflected values at positive even integers.

Are you comfortable with how the Bernoulli numbers show up for zeta evaluated at positive even integers?