Theorem: Sum of Bernoulli Numbers by Binomial Coefficients Vanishes

:$\forall n \in \Z_{>1}: \ds \sum_{k \mathop = 0}^{n - 1} \binom n k B_k = 0$
where $B_k$ denotes the $k$th Bernoulli number.

Proof:
Take the definition of Bernoulli numbers:
:$\ds \frac x {e^x - 1} = \sum_{n \mathop = 0}^\infty \frac {B_n x^n} {n!}$
From the definition of the exponential function:
{{begin-eqn}}
{{eqn | l = e^x
      | r = \sum_{n \mathop = 0}^\infty \frac {x^n} {n!}
      | c = 
}}
{{eqn | r = 1 + \sum_{n \mathop = 1}^\infty \frac {x^n} {n!}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \frac {e^x - 1} x
      | r = \sum_{n \mathop = 1}^\infty \frac {x^{n - 1} } {n!}
      | c = 
}}
{{eqn | r = 1 + \frac x {2!} + \frac {x^2} {3!} + \cdots
      | c = 
}}
{{end-eqn}}
Thus:
{{begin-eqn}}
{{eqn | l = 1
      | r = \paren {\frac x {e^x - 1} } \paren {\frac {e^x - 1} x}
      | c = 
}}
{{eqn | r = \paren {\sum_{n \mathop = 0}^\infty \frac {B_n x^n} {n!} } \paren {\sum_{n \mathop = 1}^\infty \frac {x^{n - 1} } {n!} }
      | c = 
}}
{{eqn | r = \paren {\sum_{n \mathop = 0}^\infty \frac {B_n x^n} {n!} } \paren {\sum_{n \mathop = 0}^\infty \frac {x^n} {\paren {n + 1}!} }
      | c = as both series start at zero
}}
{{end-eqn}}
By Product of Absolutely Convergent Series, we will let:
{{begin-eqn}}
{{eqn | l = a_n
      | r = \frac {B_n x^n} {n!}
      | c = 
}}
{{eqn | l = b_n
      | r = \frac {x^n} {\paren {n + 1}!} 
      | c = 
}}
{{end-eqn}}
Then:
{{begin-eqn}}
{{eqn | l = \sum_{n \mathop = 0}^\infty c_n
      | r = \paren {\sum_{n \mathop = 0}^\infty a_n} \paren {\sum_{n \mathop = 0}^\infty b_n}
      | rr= =1
      | c = 
}}
{{eqn | l = c_n
      | r = \sum_{k \mathop = 0}^n a_k b_{n - k}
      | c = 
}}
{{eqn | l = c_0
      | r = \frac {B_0 x^0} {0!} \frac {x^0} {\paren {0 + 1}!} 
      | rr= = 1
      | c = as $c_0 = \paren {a_0} \paren {b_{0 - 0} } = \paren {a_0} \paren {b_0}$
}}
{{eqn | ll= \leadsto
      | l = \sum_{n \mathop = 1}^\infty c_n
      | r = \paren {\sum_{n \mathop = 0}^\infty a_n} \paren {\sum_{n \mathop = 0}^\infty b_n} - a_0 b_0
      | rr= = 0
      | c = subtracting $1$ from both sides
}}
{{eqn | r = c_1 x + c_2 x^2 + c_3 x^3 + \cdots
      | rr= = 0
}}
{{eqn | ll= \leadsto
      | q = \forall n \in \Z_{>0}
      | l = c_n
      | r = 0
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = c_1
      | r = \frac {B_0 x^0} {0!} \frac {x^{1} } {\paren {1 + 1 }!} + \frac {B_1 x^1} {1!} \frac {x^{0} } {\paren {0 + 1 }!}
      | rr= = 0
      | rrr= = a_0 b_1 + a_1 b_0
}}
{{eqn | l = c_2
      | r = \frac {B_0 x^0} {0!} \frac {x^{2} } {\paren {2 + 1 }!} + \frac {B_1 x^1} {1!} \frac {x^{1} } {\paren {1 + 1 }!} + \frac {B_2 x^2} {2!} \frac {x^{0} } {\paren {0 + 1 }!}
      | rr= = 0
      | rrr= = a_0 b_2 + a_1 b_1 + a_2 b_0
}}
{{eqn | l = \cdots
      | r = \cdots
      | rr= = 0
}}
{{eqn | l = c_n
      | r = \frac {B_0 x^0} {0!} \frac {x^{n} } {\paren {n + 1 }!} + \frac {B_1 x^1} {1!} \frac {x^{n-1} } {\paren {n - 1 + 1 }!} + \cdots + \frac {B_n x^n} {n!} \frac {x^{0} } {\paren {0 + 1 }!}
      | rr= = 0
      | rrr= = a_0 b_n + a_1 b_{n - 1 } + a_2 b_{n - 2 } + \cdots + a_n b_0
}}
{{end-eqn}}
Multiplying $c_n$ through by $\paren {n + 1 }!$ gives:
{{begin-eqn}}
{{eqn | l = \paren {n + 1 }! c_n
      | r = \frac {B_0 x^0} {0!} \frac {\paren {n + 1 }! x^n } {\paren {n + 1 }!} + \frac {B_1 x^1} {1!} \frac {\paren {n + 1 }! x^{n-1} } {\paren {n - 1 + 1 }!} + \cdots + \frac {B_n x^n} {n!} \frac {\paren {n + 1 }! x^{0} } {\paren {0 + 1 }!}
      | rr= = 0
      | c = 
}}
{{eqn | r = x^n \paren {\frac {\paren {n + 1 }! } {0! \paren {n + 1 }!} B_0 + \frac {\paren {n + 1 }! } {1! \paren {n - 1 + 1 }!} B_1 + \cdots + \frac {\paren {n + 1 }! } {n! \paren {0 + 1 }!} B_n }
      | rr= = 0
      | c = factoring out $x^n$
}}
{{end-eqn}}
But those coefficients are the binomial coefficients:
{{begin-eqn}}
{{eqn | l = \paren {n + 1 }! c_n
      | r = \dbinom {n + 1 } 0 B_0 + \dbinom {n + 1 } 1 B_1 + \dbinom {n + 1 } 2 B_2 + \cdots + \dbinom {n + 1 } n B_n
      | rr= = 0
      | c = 
}}
{{eqn | l = n! c_{n-1 }
      | r = \dbinom n 0 B_0 + \dbinom n 1 B_1 + \dbinom n 2 B_2 + \cdots + \dbinom n {n - 1} B_{n - 1}
      | rr= = 0
      | c = 
}}
{{end-eqn}}
Hence the result.
{{qed}}