Theorem: Odd Bernoulli Numbers Vanish

Let $B_n$ denote the $n$th Bernoulli Number.
Then:
:$B_{2n + 1} = 0$
for $n \ge 1$.

Proof:
By definition, the Bernoulli numbers are given by:
:$\ds \frac x {e^x - 1} = \sum_{n \mathop = 0}^\infty \frac {B_n x^n} {n!}$
We have:
{{begin-eqn}}
{{eqn | l = \frac x {e^x - 1}
      | r = \frac x 2 \paren {\frac 2 {e^x -  1} }
      | c = multiplying top and bottom by $2$
}}
{{eqn | r = \frac x 2 \paren {\frac {e^x - e^x + 2} {e^x -  1} }
      | c = adding zero
}}
{{eqn | r = \frac x 2 \paren {\frac {\paren {e^x + 1} - \paren {e^x - 1} } {e^x -  1} }
      | c = 
}}
{{eqn | r = \frac x 2 \paren {\frac {e^x + 1} {e^x - 1} - 1}
      | c = 
}}
{{eqn | r = -\frac x 2 + \frac x 2 \paren {\frac {e^x + 1} {e^x - 1} }
      | c = 
}}
{{end-eqn}}
Take $\map f x := \dfrac x 2 \paren {\dfrac {e^x + 1} {e^x - 1} }$, and note that:
{{begin-eqn}}
{{eqn | l = \map f {-x}
      | r = \frac {-x} 2 \paren {\dfrac {e^{-x} + 1} {e^{-x} - 1} }
      | c = 
}}
{{eqn | r = -\frac {-x} 2 \paren {\dfrac {1 + e^x} {1 - e^x} }
      | c = multiplying top and bottom by $e^x$
}}
{{eqn | r = -\frac {-x} 2 \paren {\dfrac {e^x + 1} {-\paren {e^x - 1} } }
      | c = 
}}
{{eqn | r = \frac x 2 \paren {\frac {e^x + 1} {e^x - 1} }
      | c = 
}}
{{eqn | r = \map f x
      | c = 
}}
{{end-eqn}}
and so $\map f x := \dfrac x 2 \paren {\dfrac {e^x + 1} {e^x - 1} }$ is an even function.
Rewriting the definition of Bernoulli numbers
{{begin-eqn}}
{{eqn | l = \frac x {e^x - 1}
      | r = \sum_{n \mathop = 0}^\infty \frac {B_n x^n} {n!}
      | c = 
}}
{{eqn | l = \frac x {e^x - 1}
      | r = 1 - \dfrac 1 2 x + \sum_{n \mathop = 2}^\infty \frac {B_n x^n} {n!}
      | c = 
}}
{{eqn | l = \frac x {e^x - 1} + \dfrac 1 2 x
      | r = 1 + \sum_{n \mathop = 2}^\infty \frac {B_n x^n} {n!}
      | c = adding $\dfrac 1 2 x$ to both sides
}}
{{eqn | l = \frac {2x + x\paren {e^x - 1 } } {2\paren {e^x - 1 } }
      | r = 1 + \sum_{n \mathop = 2}^\infty \frac {B_n x^n} {n!}
      | c = 
}}
{{eqn | l = \dfrac x 2 \paren {\dfrac {e^x + 1} {e^x - 1} }
      | r = 1 + \sum_{n \mathop = 2}^\infty \frac {B_n x^n} {n!}
      | c = 
}}
{{eqn | l = \map f x 
      | r = 1 + \sum_{n \mathop = 2}^\infty \frac {B_n} {n!} x^n
      | c = 
}}
{{end-eqn}}
We now have:
{{begin-eqn}}
{{eqn | l = \map f x 
      | r = \map f {-x}
      | c = $\map f x$ established to be an even function above
}}
{{eqn | l = 1 + \sum_{n \mathop = 2}^\infty \frac {B_n} {n!} \paren x^n
      | r = 1 + \sum_{n \mathop = 2}^\infty \frac {B_n} {n!} \paren {-x}^n
      | c = 
}}
{{end-eqn}}
If we set $n$ to be an odd integer where $n > 1$, we have:
{{begin-eqn}}
{{eqn | l = \frac {B_{2 k + 1} } { \paren {2 k + 1}!} \paren x^{2 k + 1}
      | r = \frac {B_{2 k + 1} } { \paren {2 k + 1}!} \paren {-x}^{2 k + 1}
      | c = 
}}
{{eqn | l = B_{2 k + 1} \paren x
      | r = B_{2 k + 1} \paren {-x}
      | c = multiplying both sides by $\dfrac {\paren {2 k + 1}!} {x^2}$
}}
{{eqn | l = B_{2 k + 1}
      | r = -B_{2 k + 1}
      | c = dividing both sides by $x$
}}
{{eqn | l = 2 B_{2 k + 1}
      | r = 0
      | c = 
}}
{{eqn | l = B_{2 k + 1}
      | r = 0
      | c = 
}}
{{end-eqn}}
Therefore: $\forall n \in \N: n \ge 1$:
:$B_{2 n + 1} = 0$
{{qed}}